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N"#< "$&(*,,0246=C;=CO 0Q; 079 &*A$# >?DExb$ItS5T[Y$$b$/71q6aTIY$$$b$i ƌE7 Έ)_$$$$R$2KwsR$R$[ JY囲yĠ(H1$$$$$$$R$+GsmjKOlz$$R$0$~r>\k}s 0e0e     A@f A5% 8c8c     ?1 d0u0@Ty2 NP'p<'pA)BCD|E||S" J /@3q$ ʚ;^ :ʚ;g4dddd0Fnppp@ <4!d!d 0L46<4dddd 0L46 2*___PPT9 z? %\`The Physics ofHElectric VehiclesH}Circle Calibrate your video screen or projector. The next slide must show as a circle for the pictures to have the correct aspect ratio.|zThe Physics ofH>Electric Vehicles By J. Russell Lemon Lemon.J.Russell@ATT.netZ?H( iHow they work & And why&fxElectric Vehicles0yWhat follows is a discussion of how chemical energy is converted into electric energy and then into mechanical energy to propel a vehicle. The discussion includes how mechanical energy is used to overcome the vehicle losses of tire and aerodynamic drag, and yet have enough energy left over to climb hills and accelerate the vehicle. To go fast and far, minimize your losses.&zPf*yWilliam Thomson$ [aka Lord Kelvin] 1824-1907  When you can measure what you are speaking about, and express it in numbers, you know something about it; but when you cannot measure it, when you cannot express it in numbers, your knowledge is of a meager and unsatisfactory kind: it may be the beginning of knowledge, but you have scarcely, in your thoughts, advanced to the stage of science. V+^f{ Basic Units 3MA decimeter is a tenth of a meter, or about 3 & 15/16 inches A cubic decimeter is a liter A liter of cold water has about 1 kg of mass In San Diego that 1 kg of mass has a weight of about 9.8 newtons (force) Raise it up 1 meter and you have done 9.8 joules of work. Raise it up 1 meter in one second requires a power of 9.8 watts. VNZ0},ICEnergy3energy = force x distance joules = newtons x meters joules = volts x coulombs 1 kW-hr = 3.6 MJ [megajoules] 1 hp-hr about 2.7 MJ 1 BTU about 1054.8 J $,#7 .UPower3power = force x speed watts = newtons x meters/second watts = volts x amps one horsepower = 746 watts one horsepower = lbf x mph / 375 (i.e. 1 hp = 5 lbf @ 75 mph) 1 hp = ft-lb x rpm / 5252 >R&Rolling Resistance03`Rolling resistance of an average Radial Ply Passenger Tire inflated to 32 psi is about 1% of the weight on the tire. Rolling resistance of an average Bias Ply Tire can be more than double that of a radial ply tire with the same load and pressure. Rolling resistance is measured at maximum inflation pressure and increases as tire pressure decreases. aZaf,JRolling Resistance 203For a vehicle weighing 4000 lb, a rolling resistance of 1% of load represents a drag of 40 lb. At 60 mph, a drag of 40 lb represents a loss of 6.4 horsepower or about 4.8 kW. There are now special low rolling resistance passenger tires with a rolling resistance as low as 0.6% of load. fAir Resistance03CAir Resistance is proportional to the density of the air, the drag coefficient of the vehicle, the frontal area of the vehicle, and the speed of the vehicle squared. Typical Coefficient of Drag (Cd) for a modern passenger vehicle [with windows rolled up] is about 0.4. The EV1 was about .19. The Aptera is about .11 DDf, Air Resistance 203For a vehicle with a frontal area of 30 ft2, traveling at 60 mph at sea level with a drag coefficient of 0.4, the drag would be about 110 lb. That would be about 17.7 horsepower or about 13.2 kW. B*f ffkAir Resistance 303Power needed to overcome air resistance increases with the cube of the vehicles velocity. Going from 60 to 75 mph is an air resistance power increase of 95% Energy to overcome air resistance to go a fixed distance, increases with the square of the vehicles velocity. (GPM or KW-hr)fjVery High DragWCd between 0.4 & 0.5?Y Cd about 0.19 Z Cd about 0.11  Climbing Hills03The maximum freeway grade is 6% Some San Diego roads have grades as high as 24%. The force needed for a 4000 lb vehicle to climb a 6% grade is 240 lb. To climb a 6% grade at 60 mph, A 4000 lb vehicle needs an additional 38.4 horsepower or about 29 kW more. f  Acceleration 03If you drop something, it will accelerate at the rate of about 22 mph/sec. This is know as a 1 g acceleration. An horizontal acceleration of half that, or about 10 mph/sec would be an  aggressive , typical of a sports car with a very fast driver. An acceleration of 2 mph/sec would be  conservative , typical of an older driver, or of a Honda Civic or VW GT. lZlf Acceleration 203An acceleration of 2.2 mph/sec, or 0.1 g, of a 4000 lb vehicle would require a force of 400 lb. At 60 mph, this would require an additional 64 horsepower or about 48 kW more. $f Losses03iRoughly 4.8 + 13.2 or 18.0 kW would be needed to maintain 60 mph on a level road with a 4000 lb vehicle with typical radial tires and a cross section of 30 ft2 with a Cd of 0.4. Roughly 18.0 + 28.6 or 46.6 kW would be needed to maintain 60 mph up a 6% grade. Roughly 47 + 36 or 83 kW would be needed to accelerate at 2.2 mph/sec up the 6% grade at 60 mph. FiZZf ff Losses 2 03Running 18 kW for 40 minutes run would be 12 kW-hr of energy for a distance of 40 miles at 60 mph. With a battery pack of 144 volts, this would be about 90 amp-hr of usage. For long life of a Lead-Acid battery, the depth of discharge should be less then 80%. Even an 80% DOD would shorten the life. A 100% DOD would give a very short life. Thus the need for at least a 120 amp-hr battery for the described vehicle. fZNB<Measure Losses03[It takes a force equal to the weight of the vehicle to cause a 1 g deceleration A 1 g deceleration is about 22 mph/sec Measure how long it takes on a level road to coast from 65 to 55 mph in sec (t) Deceleration (d) = (65-55)/t mph/sec Force (f) is vehicle weight * d/22 lbs Loss is about f * 60 /375 horsepower ( 1 hp = 375 lb-mph = 746 watts )&\[ff Battery03~The source of energy for an electric vehicles is its battery. The battery must supply enough current to the electric motor in order for it to supply the needed torque. The battery must have enough voltage to force the needed current through the electric motor for the desired speed. The battery must have enough energy to supply the needed power for the needed amount of time. (~ZZf  Battery 2 03"U. S. Battery makes an 8 volt battery with a 75 amp discharge time of 85 minutes called the US-8VGC. It has a weight of about 65 lb. 18 batteries in series will supply 144 volts. 18 batteries will weight about 1170 lb. Amp-Hr rating of about 106 min @ 75 A. (178 amp-hr @ 20 hr rate) #"f"  Drive Train 03The electric motor must have enough torque to overcome the losses, climb hills and accelerate the vehicle to a useful speed. The electric motor must have enough speed for the vehicle. Gears are used to match the electric motor characteristics to the vehicle requirements. $f Drive Train 203Selected tire size is P185/60R14 Tire will make 888 revolutions per mile Each tire will hold 1047 lb at 35 psi Total gear ratio is 3.75:1 Motor RPM @ 60 mph is 3330 Maximum gross vehicle weight (including 143 lb motor, 1134 lb of batteries, 50 lb of controller & wiring, two 250 lb occupants and 250 lb of  stuff ) is 3700 lb. BKZZZZNfmElectric Motor03QSeries wound direct current motor In any gear, speed is proportional to RPM Constant torque for even acceleration Torque roughly proportional to current Increasing voltage is necessary to maintain current to maintain torque as vehicle speed and motor RPM increase Batteries must have enough voltage and current to maintain desired speed $QRfA;Electric Motor 203The selected electric motor is the Advanced DC FB1-4001 Diameter is 9.1 Weight is 143 lb Max continuous rated current is 180 A Max 1 hour rated current is 200 A Max 5 minute rated current is 340 A Current is limited by motor temperature Motor speed should be kept under 6000 rpm [High rpm causes rapid brush and bearing wear.]$MLf^Motor Characteristics(03{Torque increases with current. Back voltage increases with current and motor speed [rpm]. [Motors are also a generator].\| f$/$ff3fvVehicle Characteristics03 You select with your foot the current sent to the electric motor. With a constant current you have a constant torque. As the vehicle accelerates from a stop, the controller increases the voltage on the motor to maintain that current until there is no more voltage. [battery voltage reached] As the vehicle continues to accelerate, current and therefore torque decrease, causing acceleration to also decrease until torque is just enough to match losses and you maintain a constant speed. fwVehicle Characteristics 203k In the following graph, for a given foot setting, you follow a constant torque line up to the battery voltage and then follow a horizontal line to the right as rpm and vehicle speed increase. Note the corresponding decrease in torque. You must have enough battery voltage to push the current you need to get the torque you need to go the speed you need. (lfyjul Assumptions* 030Battery voltage is 144 volts. Maximum controller current is 500 amps. Motor is Advanced DC FB1. Vehicle gross weight is 4000 lb. Tire drag is 1% of vehicle weight. Aerodynamic Cd is 0.4. Cross sectional area is 30 ft2. Vehicle is at Sea Level.`ff ffC{FD>Warning0  Note that the highest force in the previous slide is for a current of almost 500 A that will quickly overheat the motor. The continuous current must be less then 180 A and that means that the continuous force must be less than 1/3 of the maximum force shown.   ~ Motor Comment03 Remember that power is the product of torque and rpm. With the ADC FB1-4001, the 200 A continuous rating is a torque limit of about 30 ft-lbs. At 30 ft-lbs, it takes about 144 V for a motor speed of 5500 rpm. This is about 31 hp. Actually, I2R losses in battery, controller and wiring will reduce the actual voltage available to the motor. At 80% DOD with a 200 A load, the maximum voltage at the motor may be as low as 120 V for only 4500 rpm. [25 hp]8f ffMotor Comment 203 Gearing the motor for 4500 rpm at the top vehicle speed [70 mph?] will take full advantage of the capability of the battery, controller and motor in the real world. Too many car conversions fail to take into account worst case conditions. [The last hill to climb with batteries at 80% DOD.] Of course some have the option to shift to a lower gear and struggle at a lower speed. ftMeasure Performance*030xIt takes a force equal to the weight of the vehicle in addition to the force to over-come losses to cause a 1 g acceleration. Measure how long it takes on a level road to accelerate from 55 to 65 mph in seconds (t). Acceleration (a) at 60 mph is about 10/t. Force (f) is about weight * a/22 lbs. Acceleration Hp is about f * 60 / 375. Total Hp is Acceleration Hp + Loss Hp. 6xZZwf)%Range*0304Now that we have a rough idea of the vehicle s performance, the next question is how far will it go on a charge. In other words, what is its range? Range should really be determined by how far it will go on 80% of a charge since completely discharging a battery will ruin it. Note that the capacity [amp-hr] decreases as the current increases. Also note that the voltage decreases as the charge is used up.(ffRange 2*030To estimate range at a given speed, determine the force needed at that speed. The force (lb) x speed (mph) / 375 is the hp needed to maintain that speed. Multiply by .746 to get kW. Divide by the battery voltage to get battery current. Estimate battery amp-hr at that current and divide by the current. Multiply by 0.8 to get the approximate number of hours. Multiply by the speed to get an estimate of range. (ff+'Available Current*030\The total capacity of the battery is non-linear. The minutes the battery can provide power decreases faster then the amps supplied by the US 8V GC battery: 1041 minutes @ 10 amps 341 minutes @ 25 amps 146 minutes @ 50 amps 94 minutes @ 75 amps 66 minutes @ 100 amps 50 minutes @ 125 amps(]ffb83NOTICE03 The numbers used on the previous slides were taken from the best information and estimates available. Exact measured numbers were not available. Therefore, notice is given that the conclusions are approximate ballpark estimates. Actual performance to be determined."  f94Charging 0* U.S. Battery recommends that: Voltage not exceed 2.585 V per cell Current not exceed AH/10 Time not exceed 10 hours http://www.usbattery.com/pages/usbspecs.htm In other words, for a  144 volt pack, the charging current should not exceed 165/10 or about 16.5 amps until limited by the total voltage that must not exceed 186 volts. Maximum charge time is 10 hours. Check water level after charge.  %$fY+f>  :5Charging 0m186 volts times 16.5 amps is 3065 watts. That would be about 26 amps from a 120 volt source, or 13 amps from a 240 volts source, not taking into account efficiency of the charger. If time were short, the batteries could be charged at 25 amps. That would be 4650 watts, or almost 20 amps from a 240 volt source. A 30 amp 240 volt service is best for charging. \m$jfk<7. . . ie Hybrid0CFor long trips a small motor-generator can be added to extend range. Motor generators are made to run on a variety of different fuels. Commercial motor-generators include gasoline, diesel, propane, etc. Be sure the controller can take the higher voltage. Voltage should not exceed the maximum battery charging voltage. >D$B(|K f~San Diego Car Conversion Project July-August 2008 Physics of Electric Vehicles by J. Russell Lemon Lemon.J.Russell@ATT.net f4f(J J fff"f 0f}aThe End`  To Return http://EVAoSD.com $ 0/,KLMNOPQ R S T VWYZ[\h j"k#o'q)r*t,z014;@A ` ` ̙33` 333MMM` ff3333f` f` f` ff333f>?" dd@,|?" dd@   " @ ` n?" dd@   @@``PR    @ ` ` p>> (    6hp P  T Click to edit Master title style! !  0s   RClick to edit Master text styles Second level Third level Fourth level Fifth level!     S  0w ``  X*  0h} `   Z*  0< `   Z*B  s *޽h ? ff333f Default Design 0 p*(    0 P    X*  33   0     Z*( 33 d  c $ ?    0  @  RClick to edit Master text styles Second level Third level Fourth level Fifth level!     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The next slide must show as a circle for the pictures to have the correct aspect ratio.|zThe Physics ofH6Electric Vehicles By Russ Lemon Russ@FarTooMuch.Info Z7H(  "!  0!5iHow they work & And why&fxElectric Vehicles0yWhat follows is a discussion of how chemical energy is converted into electric energy and then into mechanical energy to propel a vehicle. The discussion includes how mechanical energy is used to overcome the vehicle losses of tire and aerodynamic drag, and yet have enough energy left over to climb hills and accelerate the vehicle. To go fast and far, minimize your losses.&zPf*yWilliam Thomson$ [aka Lord Kelvin] 1824-1907  When yoRoot EntrydO)@ocu PicturesCurrent User,SummaryInformation(  !"#$%&'()*+,-./0123456_8RussRussOh+'0 `h  Electric VehicleseJ. Russell LemoneRusssse100Microsoft PowerPointP@ ;@`-,k@PC\u Gg  S!& &&#TNPP2OM 0DTimes New RomanU|d0|Wo 0DComic Sans MSnU|d0|Wo 0Bg33(.  @n?" dd@  @@`` tV  *,44+6-O!"=!#%')++/357>D<>DP  /R< /8: %)B#"   !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~      !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~      !"#$%&'()*+,./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~?@T EF01UVxb$ItS5T[Y $ $ b$/71q6aTIY $ $ $ b$i ƌE7 Έ)_ $ $ $ $ R$2KwsR $ R$[ JY囲yĠ(H1 $ $ $ $ $ $PowerPoint Document(-\DocumentSummaryInformation8p $ R$+GsmjKOlz $ $ R$0$~r>\k}s  0e0e     A@f A5% 8c8c     ?1 d0u0@Ty2 NP'p<'pA)BCD|E||S" J /@3q$ ʚ;^ :ʚ;g4dddd0pppp@ <4!d!d 0,V<4dddd 0,V 2*___PPT9 z? %e``The Physics ofHElectric VehiclesH}Circle Calibrate your video screen or projector. The next slide must show as a circle for the pictures to have the correct aspect ratio.|zThe Physics ofH6Electric Vehicles By Russ Lemon Russ@FarTooMuch.Info Z7H(  "!  0!5iHow they work & And why&fxElectric Vehicles0yWhat follows is a discussion of how chemical energy is converted into electric energy and then into mechanical energy to propel a vehicle. The discussion includes how mechanical energy is used to overcome the vehicle losses of tire and aerodynamic drag, and yet have enough energy left over to climb hills and accelerate the vehicle. To go fast and far, minimize your losses.&zPf*yWilliam Thomson$ [aka Lord Kelvin] 1824-1907  When you can measure what you are speaking about, and express it in numbers, you know something about it; but when you cannot measure it, when you cannot express it in numbers, your knowledge is of a meager and unsatisfactory kind: it may be the beginning of knowledge, but you have scarcely, in your thoughts, advanced to the stage of science. V+^f{ Basic Units 3MA decimeter is a tenth of a meter, or about 3 & 15/16 inches A cubic decimeter is a liter A liter of cold water has about 1 kg of mass In San Diego that 1 kg of mass has a weight of about 9.8 newtons (force) Raise it up 1 meter and you have done 9.8 joules of work. Raise it up 1 meter in one second requires a power of 9.8 watts. VNZ0},ICEnergy3energy = force x distance joules = newtons x meters joules = volts x coulombs 1 kW-hr = 3.6 MJ [megajoules] 1 hp-hr about 2.7 MJ 1 BTU about 1054.8 J $,#7 .UPower3power = force x speed watts = newtons x meters/second watts = volts x amps one horsepower = 746 watts one horsepower = lbf x mph / 375 (i.e. 1 hp = 5 lbf @ 75 mph) 1 hp = ft-lb x rpm / 5252 >R&Rolling Resistance03`Rolling resistance of an average Radial Ply Passenger Tire inflated to 32 psi is about 1% of the weight on the tire. Rolling resistance of an average Bias Ply Tire can be more than double that of a radial ply tire with the same load and pressure. Rolling resistance is measured at maximum inflation pressure and increases as tire pressure decreases. aZaf,JRolling Resistance 203For a vehicle weighing 4000 lb, a rolling resistance of 1% of load represents a drag of 40 lb. At 60 mph, a drag of 40 lb represents a loss of 6.4 horsepower or about 4.8 kW. There are now special low rolling resistance passenger tires with a rolling resistance as low as 0.6% of load. fAir Resistance03CAir Resistance is proportional to the density of the air, the drag coefficient of the vehicle, the frontal area of the vehicle, and the speed of the vehicle squared. Typical Coefficient of Drag (Cd) for a modern passenger vehicle [with windows rolled up] is about 0.4. The EV1 was about .19. The Aptera is about .11 DDf, Air Resistance 203For a vehicle with a frontal area of 30 ft2, traveling at 60 mph at sea level with a drag coefficient of 0.4, the drag would be about 110 lb. That would be about 17.7 horsepower or about 13.2 kW. B*f ffkAir Resistance 303Power needed to overcome air resistance increases with the cube of the vehicles velocity. Going from 60 to 75 mph is an air resistance power increase of 95% Energy to overcome air resistance to go a fixed distance, increases with the square of the vehicles velocity. (GPM or KW-hr)fjVery High DragWCd between 0.4 & 0.5?Y Cd about 0.19 Z Cd about 0.11  Climbing Hills03The maximum freeway grade is 6% Some San Diego roads have grades as high as 24%. The force needed for a 4000 lb vehicle to climb a 6% grade is 240 lb. To climb a 6% grade at 60 mph, A 4000 lb vehicle needs an additional 38.4 horsepower or about 29 kW more. f  Acceleration 03If you drop something, it will accelerate at the rate of about 22 mph/sec. This is know as a 1 g acceleration. An horizontal acceleration of half that, or about 10 mph/sec would be an  aggressive , typical of a sports car with a very fast driver. An acceleration of 2 mph/sec would be  conservative , typical of an older driver, or of a Honda Civic or VW GT. lZlf Acceleration 203An acceleration of 2.2 mph/sec, or 0.1 g, of a 4000 lb vehicle would require a force of 400 lb. At 60 mph, this would require an additional 64 horsepower or about 48 kW more. $f Losses03iRoughly 4.8 + 13.2 or 18.0 kW would be needed to maintain 60 mph on a level road with a 4000 lb vehicle with typical radial tires and a cross section of 30 ft2 with a Cd of 0.4. Roughly 18.0 + 28.6 or 46.6 kW would be needed to maintain 60 mph up a 6% grade. Roughly 47 + 36 or 83 kW would be needed to accelerate at 2.2 mph/sec up the 6% grade at 60 mph. FiZZf ff Losses 2 03Running 18 kW for 40 minutes run would be 12 kW-hr of energy for a distance of 40 miles at 60 mph. With a battery pack of 144 volts, this would be about 90 amp-hr of usage. For long life of a Lead-Acid battery, the depth of discharge should be less then 80%. Even an 80% DOD would shorten the life. A 100% DOD would give a very short life. Thus the need for at least a 120 amp-hr battery for the described vehicle. fZNB<Measure Losses03[It takes a force equal to the weight of the vehicle to cause a 1 g deceleration A 1 g deceleration is about 22 mph/sec Measure how long it takes on a level road to coast from 65 to 55 mph in sec (t) Deceleration (d) = (65-55)/t mph/sec Force (f) is vehicle weight * d/22 lbs Loss is about f * 60 /375 horsepower ( 1 hp = 375 lb-mph = 746 watts )&\[ff  Battery 1 03~The source of energy for an electric vehicles is its battery. The battery must supply enough current to the electric motor in order for it to supply the needed torque. The battery must have enough voltage to force the needed current through the electric motor for the desired speed. The battery must have enough energy to supply the needed power for the needed amount of time. (~ZZf  Battery 2 03"U. S. Battery makes an 8 volt battery with a 75 amp discharge time of 85 minutes called the US-8VGC. It has a weight of about 65 lb. 18 batteries in series will supply 144 volts. 18 batteries will weight about 1170 lb. Amp-Hr rating of about 106 min @ 75 A. (178 amp-hr @ 20 hr rate) #"f"8Battery 3  [rules of thumb]* 03$3Lead-acid batteries in an electric vehicle need to be at least 33% of a good vehicles gross weight to get a range of more than 40 miles with conservative driving. To get good performance, you need at least 33% of the vehicles gross weight to be active, on-line battery. $fJBattery 4  [lead-acid battery life]*& 03 3Do not exceed 80% depth of discharge. Keep battery voltage within normal range. [For 144 V pack, keep pack above 120 V and below 185 V at all times.] Limit maximum current. [Excessive current leads to short life and even battery failure.] [ $f  Drive Train 03The electric motor must have enough torque to overcome the losses, climb hills and accelerate the vehicle to a useful speed. The electric motor must have enough speed for the vehicle. Gears are used to match the electric motor characteristics to the vehicle requirements. $f Drive Train 203Selected tire size is P185/60R14 Tire will make 888 revolutions per mile Each tire will hold 1047 lb at 35 psi Total gear ratio is 3.75:1 Motor RPM @ 60 mph is 3330 Maximum gross vehicle weight (including 143 lb motor, 1134 lb of batteries, 50 lb of controller & wiring, two 250 lb occupants and 250 lb of  stuff ) is 3700 lb. BKZZZZNf$mElectric Motor03QSeries wound direct current motor In any gear, speed is proportional to RPM Constant torque for even acceleration Torque roughly proportional to current Increasing voltage is necessary to maintain current to maintain torque as vehicle speed and motor RPM increase Batteries must have enough voltage and current to maintain desired speed $QRfA;Electric Motor 203The selected electric motor is the Advanced DC FB1-4001 Diameter is 9.1 Weight is 143 lb Max continuous rated current is 180 A Max 1 hour rated current is 200 A Max 5 minute rated current is 340 A Current is limited by motor temperature Motor speed should be kept under 6000 rpm [High rpm causes rapid brush and bearing wear.]$MLf^Motor Characteristics(03{Torque increases with current. Back voltage increases with current and motor speed [rpm]. [Motors are also a generator].\| f$/$ff3fvVehicle Characteristics03 You select with your foot the current sent to the electric motor. With a constant current you have a constant torque. As the vehicle accelerates from a stop, the controller increases the voltage on the motor to maintain that current until there is no more voltage. [battery voltage reached] As the vehicle continues to accelerate, current and therefore torque decrease, causing acceleration to also decrease until torque is just enough to match losses and you maintain a constant speed. fwVehicle Characteristics 203k In the following graph, for a given foot setting, you follow a constant torque line up to the battery voltage and then follow a horizontal line to the right as rpm and vehicle speed increase. Note the corresponding decrease in torque. You must have enough battery voltage to push the current you need to get the torque you need to go the speed you need. (lfyjul Assumptions* 030Battery voltage is 144 volts. Maximum controller current is 500       !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz|}~amps. Motor is Advanced DC FB1. Vehicle gross weight is 4000 lb. Tire drag is 1% of vehicle weight. Aerodynamic Cd is 0.4. Cross sectional area is 30 ft2. Vehicle is at Sea Level.`ff ffC{FD>Warning0  Note that the highest force in the previous slide is for a current of almost 500 A that will quickly overheat the motor. The continuous current must be less then 180 A and that means that the continuous force must be less than 1/3 of the maximum force shown.   ~ Motor Comment03 Remember that power is the product of torque and rpm. With the ADC FB1-4001, the 200 A continuous rating is a torque limit of about 30 ft-lbs. At 30 ft-lbs, it takes about 144 V for a motor speed of 5500 rpm. This is about 31 hp. Actually, I2R losses in battery, controller and wiring will reduce the actual voltage available to the motor. At 80% DOD with a 200 A load, the maximum voltage at the motor may be as low as 120 V for only 4500 rpm. [25 hp]8f ffMotor Comment 203 Gearing the motor for 4500 rpm at the top vehicle speed [70 mph?] will take full advantage of the capability of the battery, controller and motor in the real world. Too many car conversions fail to take into account worst case conditions. [The last hill to climb with batteries at 80% DOD.] Of course some have the option to shift to a lower gear and struggle at a lower speed. ftMeasure Performance*030xIt takes a force equal to the weight of the vehicle in addition to the force to over-come losses to cause a 1 g acceleration. Measure how long it takes on a level road to accelerate from 55 to 65 mph in seconds (t). Acceleration (a) at 60 mph is about 10/t. Force (f) is about weight * a/22 lbs. Acceleration Hp is about f * 60 / 375. Total Hp is Acceleration Hp + Loss Hp. 6xZZwf)%Range*0304Now that we have a rough idea of the vehicle s performance, the next question is how far will it go on a charge. In other words, what is its range? Range should really be determined by how far it will go on 80% of a charge since completely discharging a battery will ruin it. Note that the capacity [amp-hr] decreases as the current increases. Also note that the voltage decreases as the charge is used up.(ffRange 2*030To estimate range at a given speed, determine the force needed at that speed. The force (lb) x speed (mph) / 375 is the hp needed to maintain that speed. Multiply by .746 to get kW. Divide by the battery voltage to get battery current. Estimate battery amp-hr at that current and divide by the current. Multiply by 0.8 to get the approximate number of hours. Multiply by the speed to get an estimate of range. (ff+'Available Current*030\The total capacity of the battery is non-linear. The minutes the battery can provide power decreases faster then the amps supplied by the US 8V GC battery: 1041 minutes @ 10 amps 341 minutes @ 25 amps 146 minutes @ 50 amps 94 minutes @ 75 amps 66 minutes @ 100 amps 50 minutes @ 125 amps(]ffb83NOTICE03 The numbers used on the previous slides were taken from the best information and estimates available. Exact measured numbers were not available. Therefore, notice is given that the conclusions are approximate ballpark estimates. Actual performance to be determined."  f94Charging 0* U.S. Battery recommends that: Voltage not exceed 2.585 V per cell Current not exceed AH/10 Time not exceed 10 hours http://www.usbattery.com/pages/usbspecs.htm In other words, for a  144 volt pack, the charging current should not exceed 165/10 or about 16.5 amps until limited by the total voltage that must not exceed 186 volts. Maximum charge time is 10 hours. Check water level after charge.  %$fY+f>  :5Charging 0m186 volts times 16.5 amps is 3065 watts. That would be about 26 amps from a 120 volt source, or 13 amps from a 240 volts source, not taking into account efficiency of the charger. If time were short, the batteries could be charged at 25 amps. That would be 4650 watts, or almost 20 amps from a 240 volt source. A 30 amp 240 volt service is best for charging. \m$jfk<7. . . ie Hybrid0CFor long trips a small motor-generator can be added to extend range. Motor generators are made to run on a variety of different fuels. Commercial motor-generators include gasoline, diesel, propane, etc. Be sure the controller can take the higher voltage. Voltage should not exceed the maximum battery charging voltage. >D$B(|K f~San Diego Car Conversion Project July-August 2008 Physics of Electric Vehicles by J. Russell Lemon Lemon.J.Russell@ATT.net f4f(J J fff"f 0f}aThe End`  To Return http://EVAoSD.com $ 0/dKLMNOPQ R S T VWYZ[\h j"k#o'q)r*t,z014;@ABC p<0(  <x < c $\`  x < c $@YP`  B < s *޽h ? ff3333f  <0(  <x < c $D,Ppp   x < c $ =0   H < 0޽h ? ff3333f B h0(  hx h c $p   x h c $c   H h 0޽h ? ff3333f C Pp0(  px p c $\jp   x p c $Z   H p 0޽h ? ff3333f 0 l((  l^ l S     l c $$g @    H l 0޽h ? ̙33 0 `t((  t^ t S     t c $J @    H t 0޽h ? ̙33r,@ՙݗ Dzw(i & TNPP &&TNPP    --- !---&G&w@ Xwaw @w f2- &Gy& --1q@-- @Times New RomanXwaw @w f2- .2 The Physics of;0+51.%+%1.--`8--  3.!2 Electric Vehicles;+* +E+/++%.--"System2 f2 !-&TNPP &՜.+,D՜.+,    ILetter Paper (8.5x11 in)oett\6  9Times New RomanComic Sans MSDefault DesignThe Physics ofCirclePu can measure what you are speaking about, and express it in numbers, you know something about it; but when you cannot measure it, when you cannot express it in numbers, your knowledge is of a meager and unsatisfactory kind: it may be the beginning of knowledge, but you have scarcely, in your thoughts, advanced to the stage of science. V+^f{ Basic Units 3MA decimeter is a tenth of a meter, or about 3 & 15/16 inches A cubic decimeter is a liter A liter of cold water has about 1 kg of mass In San Diego that 1 kg of mass has a weight of about 9.8 newtons (force) Raise it up 1 meter and you have done 9.8 joules of work. Raise it up 1 meter in one second requires a power of 9.8 watts. VNZ0},ICEnergy3energy = force x distance joules = newtons x meters joules = volts x coulombs 1 kW-hr = 3.6 MJ [megajoules] 1 hp-hr about 2.7 MJ 1 BTU about 1054.8 J $,#7 .UPower3power = force x speed watts = newtons x meters/second watts = volts x amps one horsepower = 746 watts one horsepower = lbf x mph / 375 (i.e. 1 hp = 5 lbf @ 75 mph) 1 hp = ft-lb x rpm / 5252 >R&Rolling Resistance03`Rolling resistance of an average Radial Ply Passenger Tire inflated to 32 psi is about 1% of the weight on the tire. Rolling resistance of an average Bias Ply Tire can be more than double that of a radial ply tire with the same load and pressure. Rolling resistance is measured at maximum inflation pressure and increases as tire pressure decreases. aZaf,JRolling Resistance 203For a vehicle weighing 4000 lb, a rolling resistance of 1% of load represents a drag of 40 lb. At 60 mph, a drag of 40 lb represents a loss of 6.4 horsepower or about 4.8 kW. There are now special low rolling resistance passenger tires with a rolling resistance as low as 0.6% of load. fAir Resistance03CAir Resistance is proportional to the density of the air, the drag coefficient of the vehicle, the frontal area of the vehicle, and the speed of the vehicle squared. Typical Coefficient of Drag (Cd) for a modern passenger vehicle [with windows rolled up] is about 0.4. The EV1 was about .19. The Aptera is about .11 DDf, Air Resistance 203For a vehicle with a frontal area of 30 ft2, traveling at 60 mph at sea level with a drag coefficient of 0.4, the drag would be about 110 lb. That would be about 17.7 horsepower or about 13.2 kW. B*f ffkAir Resistance 303Power needed to overcome air resistance increases with the cube of the vehicles velocity. Going from 60 to 75 mph is an air resistance power increase of 95% Energy to overcome air resistance to go a fixed distance, increases with the square of the vehicles velocity. (GPM or KW-hr)fjVery High DragWCd between 0.4 & 0.5?Y Cd about 0.19 Z Cd about 0.11  Climbing Hills03The maximum freeway grade is 6% Some San Diego roads have grades as high as 24%. The force needed for a 4000 lb vehicle to climb a 6% grade is 240 lb. To climb a 6% grade at 60 mph, A 4000 lb vehicle needs an additional 38.4 horsepower or about 29 kW more. f  Acceleration 03If you drop something, it will accelerate at the rate of about 22 mph/sec. This is know as a 1 g acceleration. An horizontal acceleration of half that, or about 10 mph/sec would be an  aggressive , typical of a sports car with a very fast driver. An acceleration of 2 mph/sec would be  conservative , typical of an older driver, or of a Honda Civic or VW GT. lZlf Acceleration 203An acceleration of 2.2 mph/sec, or 0.1 g, of a 4000 lb vehicle would require a force of 400 lb. At 60 mph, this would require an additional 64 horsepower or about 48 kW more. $f Losses03iRoughly 4.8 + 13.2 or 18.0 kW would be needed to maintain 60 mph on a level road with a 4000 lb vehicle with typical radial tires and a cross section of 30 ft2 with a Cd of 0.4. Roughly 18.0 + 28.6 or 46.6 kW would be needed to maintain 60 mph up a 6% grade. Roughly 47 + 36 or 83 kW would be needed to accelerate at 2.2 mph/sec up the 6% grade at 60 mph. FiZZf ff Losses 2 03Running 18 kW for 40 minutes run would be 12 kW-hr of energy for a distance of 40 miles at 60 mph. With a battery pack of 144 volts, this would be about 90 amp-hr of usage. For long life of a Lead-Acid battery, the depth of discharge should be less then 80%. Even an 80% DOD would shorten the life. A 100% DOD would give a very short life. Thus the need for at least a 120 amp-hr battery for the described vehicle. fZNB<Measure Losses03[It takes a force equal to the weight of the vehicle to cause a 1 g deceleration A 1 g deceleration is about 22 mph/sec Measure how long it takes on a level road to coast from 65 to 55 mph in sec (t) Deceleration (d) = (65-55)/t mph/sec Force (f) is vehicle weight * d/22 lbs Loss is about f * 60 /375 horsepower ( 1 hp = 375 lb-mph = 746 watts )&\[ff  Battery 1 03~The source of energy for an electric vehicles is its battery. The battery must supply enough current to the electric motor in order for it to supply the needed torque. The battery must have enough voltage to force the needed current through the electric motor for the desired speed. The battery must have enough energy to supply the needed power for the needed amount of time. (~ZZf  Battery 2 03"U. S. Battery makes an 8 volt battery with a 75 amp discharge time of 85 minutes called the US-8VGC. It has a weight of about 65 lb. 18 batteries in series will supply 144 volts. 18 batteries will weight about 1170 lb. Amp-Hr rating of about 106 min @ 75 A. (178 amp-hr @ 20 hr rate) #"f"8Battery 3  [rules of thumb]* 03$3Lead-acid batteries in an electric vehicle need to be at least 33% of a good vehicles gross weight to get a range of more than 40 miles with conservative driving. To get good performance, you need at least 33% of the vehicles gross weight to be active, on-line battery. $fJBattery 4  [lead-acid battery life]*& 03 3]Do not exceed 80% depth of discharge. Keep battery voltage within normal range. [For 144 V pack, keep pack above 120 V and below 185 V at all times.] Limit maximum current. [Excessive current leads to short life and even battery failure.] [Keep maximum current below the current that gives a full charge to 80% Discharge time of 20 minutes.] $]^f  Drive Train 03The electric motor must have enough torque to overcome the losses, climb hills and accelerate the vehicle to a useful speed. The electric motor must have enough speed for the vehicle. Gears are used to match the electric motor characteristics to the vehicle requirements. $f Drive Train 203Selected tire size is P185/60R14 Tire will make 888 revolutions per mile Each tire will hold 1047 lb at 35 psi Total gear ratio is 3.75:1 Motor RPM @ 60 mph is 3330 Maximum gross vehicle weight (including 143 lb motor, 1134 lb of batteries, 50 lb of controller & wiring, two 250 lb occupants and 250 lb of  stuff ) is 3700 lb. BKZZZZNf$mElectric Motor03QSeries wound direct current motor In any gear, speed is proportional to RPM Constant torque for even acceleration Torque roughly proportional to current Increasing voltage is necessary to maintain current to maintain torque as vehicle speed and motor RPM increase Batteries must have enough voltage and current to maintain desired speed $QRfA;Electric Motor 203The selected electric motor is the Advanced DC FB1-4001 Diameter is 9.1 Weight is 143 lb Max continuous rated current is 180 A Max 1 hour rated current is 200 A Max 5 minute rated current is 340 A Current is limited by motor temperature Motor speed should be kept under 6000 rpm [High rpm causes rapid brush and bearing wear.]$MLf^Motor Characteristics(03{Torque increases with current. Back voltage increases with current and motor speed [rpm]. [Motors are also a generator].\| f$/$ff3fvVehicle Characteristics03 You select with your foot the current sent to the electric motor. With a constant current you have a constant torque. As the vehicle accelerates from a stop, the controller increases the voltage on the motor to maintain that current until there is no more voltage. [battery voltage reached] As the vehicle continues to accelerate, current and therefore torque decrease, causing acceleration to also decrease until torque is just enough to match losses and you maintain a constant speed. fwVehicle Characteristics 203k In the following graph, for a given foot setting, you follow a constant torque line up to the battery voltage and then follow a horizontal line to the right as rpm and vehicle speed increase. Note the corresponding decrease in torque. You must have enough battery voltage to push the current you need to get the torque you need to go the speed you need. (lfyjul Assumptions* 030Battery voltage is 144 volts. Maximum controller current is 500 amps. Motor is Advanced DC FB1. Vehicle gross weight is 4000 lb. Tire drag is 1% of vehicle weight. Aerodynamic Cd is 0.4. Cross sectional area is 30 ft2. Vehicle is at Sea Level.`ff ffC{FD>Warning0  Note that the highest force in the previous slide is for a current of almost 500 A that will quickly overheat the motor. The continuous current must be less then 180 A and that means that the continuous force must be less than 1/3 of the maximum force shown.   ~ Motor Comment03 Remember that power is the product of torque and rpm. With the ADC FB1-4001, the 200 A continuous rating is a torque limit of about 30 ft-lbs. At 30 ft-lbs, it takes about 144 V for a motor speed of 5500 rpm. This is about 31 hp. Actually, I2R losses in battery, controller and wiring will reduce the actual voltage available to the motor. At 80% DOD with a 200 A load, the maximum voltage at the motor may be as low as 120 V for only 4500 rpm. [25 hp]8f ffMotor Comment 203 Gearing the motor for 4500 rpm at the top vehicle speed [70 mph?] will take full advantage of the capability of the battery, controller and motor in the real world. Too many car conversions fail to take into account worst case conditions. [The last hill to climb with batteries at 80% DOD.] Of course some have the option to shift to a lower gear and struggle at a lower speed. ftMeasure Performance*030xIt takes a force equal to the weight of the vehicle in addition to the force to over-come losses to cause a 1 g acceleration. Measure how long it takes on a level road to accelerate from 55 to 65 mph in seconds (t). Acceleration (a) at 60 mph is about 10/t. Force (f) is about weight * a/22 lbs. Acceleration Hp is about f * 60 / 375. Total Hp is Acceleration Hp + Loss Hp. 6xZZwf)%Range*0304Now that we have a rough idea of the vehicle s performance, the next question is how far will it go on a charge. In other words, what is its range? Range should really be determined by how far it will go on 80% of a charge since completely discharging a battery will ruin it. Note that the capacity [amp-hr] decreases as the current increases. Also note that the voltage decreases as the charge is used up.(ffRange 2*030To estimate range at a given speed, determine the force needed at that speed. The force (lb) x speed (mph) / 375 is the hp needed to maintain that speed. Multiply by .746 to get kW. Divide by the battery voltage to get battery current. Estimate battery amp-hr at that current and divide by the current. Multiply by 0.8 to get the approximate number of hours. Multiply by the speed to get an estimate of range. (ff+'Available Current*030\The total capacity of the battery is non-linear. The minutes the battery can provide power decreases faster then the amps supplied by the US 8V GC battery: 1041 minutes @ 10 amps 341 minutes @ 25 amps 146 minutes @ 50 amps 94 minutes @ 75 amps 66 minutes @ 100 amps 50 minutes @ 125 amps(]ffb83NOTICE03 The numbers used on the previous slides were taken from the best information and estimates available. Exact measured numbers were not available. Therefore, notice is given that the conclusions are approximate ballpark estimates. Actual performance to be determined."  f94Charging 0* U.S. Battery recommends that: Voltage not exceed 2.585 V per cell Current not exceed AH/10 Time not exceed 10 hours http://www.usbattery.com/pages/usbspecs.htm In other words, for a  144 volt pack, the charging current should not exceed 165/10 or about 16.5 amps until limited by the total voltage that must not exceed 186 volts. Maximum charge time is 10 hours. Check water level after charge.  %$fY+f>  :5Charging 0m186 volts times 16.5 amps is 3065 watts. That would be about 26 amps from a 120 volt source, or 13 amps from a 240 volts source, not taking into account efficiency of the charger. If time were short, the batteries could be charged at 25 amps. That would be 4650 watts, or almost 20 amps from a 240 volt source. A 30 amp 240 volt service is best for charging. \m$jfk<7. . . ie Hybrid0CFor long trips a small motor-generator can be added to extend range. Motor generators are made to run on a variety of different fuels. Commercial motor-generators include gasoline, diesel, propane, etc. Be sure the controller can take the higher voltage. Voltage should not exceed the maximum battery charging voltage. >D$B(|K f~San Diego Car Conversion Project July-August 2008 Physics of Electric Vehicles by J. Russell Lemon Lemon.J.Russell@ATT.net f4f(J J fff"f 0f}aThe End`  To Return http://EVAoSD.com $ 0/dKLMNOPQ R S T VWYZ[\h j"k#o'q)r*t,z014;@ABC C Pp0(  px p c $\jp   x p c $Z 0  H p 0޽h ? ff3333froɝgDw( 5  6"http://EVAoSD.com< (Russ@FarTooMuch.Info<(Russ@FarTooMuch.Infol/ 0DTimes New Roman`F|d0|Wo 0DComic Sans MSn`F|d0|Wo 0Bg33(.  @n?" dd@  @@`` |V  *,67I 9-Q!"?!#%'),.2579@F?AGR 2T> 2:= %)D#" ABV GH/01xb$ItS5T[Y $ $ b$/71q6aTIY $ $ $ b$i ƌE7 Έ)_ $ $ $ $ R$2KwsR $ R$[ JY囲yĠ(H1 $ $ $ $ $ $ $ R$+GsmjKOlz $ $ R$0$~r>\k}s  0e0e     A@f A5% 8c8c     ?1 d0u0@Ty2 NP'p<'pA)BCD|E||S" J /@3q$ ʚ;^ :ʚ;g4dddd0pppp@ <4!d!d 0,LG<4dddd 0,LG 2*___PPT9 z? %``The Physics ofHElectric VehiclesH}Circle Calibrate your video screen or projector. The next slide must show as a circle for the pictures to have the correct aspect ratio.|zThe Physics ofH6Electric Vehicles By Russ Lemon Russ@FarTooMuch.Info Z7H(  "!  0!5iHow they work & And why&fxElectric Vehicles0yWhat follows is a discussion of how chemical energy is converted into electric energy and then into mechanical energy to propel a vehicle. The discussion includes how mechanical energy is used to overcome the vehicle losses of tire and aerodynamic drag, and yet have enough energy left over to climb hills and accelerate the vehicle. To go fast and far, minimize your losses.&zPf*yWilliam Thomson$ [aka Lord Kelvin] 1824-1907  When you can measure what you are speaking about, and express it in numbers, you know something about it; but when you cannot measure it, when you cannot express it in numbers, your knowledge is of a meager and unsatisfactory kind: it may be the beginning of knowledge, but you have scarcely, in your thoughts, advanced to the stage of science. V+^f{ Basic Units 3MA decimeter is a tenth of a meter, or about 3 & 15/16 inches A cubic decimeter is a liter A liter of cold water has about 1 kg of mass In San Diego that 1 kg of mass has a weight of about 9.8 newtons (force) Raise it up 1 meter and you have done 9.8 joules of work. Raise it up 1 meter in one second requires a power of 9.8 watts. VNZ0},ICEnergy3energy = force x distance joules = newtons x meters joules = volts x coulombs 1 kW-hr = 3.6 MJ [megajoules] 1 hp-hr about 2.7 MJ 1 BTU about 1054.8 J $,#7 .UPower3power = force x speed watts = newtons x meters/second watts = volts x amps one horsepower = 746 watts one horsepower = lbf x mph / 375 (i.e. 1 hp = 5 lbf @ 75 mph) 1 hp = ft-lb x rpm / 5252 >R&Rolling Resistance03`Rolling resistance of an average Radial Ply Passenger Tire inflated to 32 psi is about 1% of the weight on the tire. Rolling resistance of an average Bias Ply Tire can be more than double that of a radial ply tire with the same load and pressure. Rolling resistance is measured at maximum inflation pressure and increases as tire pressure decreases. aZaf,JRolling Resistance 203For a vehicle weighing 4000 lb, a rolling resistance of 1% of load represents a drag of 40 lb. At 60 mph, a drag of 40 lb represents a loss of 6.4 horsepower or about 4.8 kW. There are now special low rolling resistance passenger tires with a rolling resistance as low as 0.6% of load. fAir Resistance03CAir Resistance is proportional to the density of the air, the drag coefficient of the vehicle, the frontal area of the vehicle, and the speed of the vehicle squared. Typical Coefficient of Drag (Cd) for a modern passenger vehicle [with windows rolled up] is about 0.4. The EV1 was about .19. The Aptera is about .11 DDf, Air Resistance 203For a vehicle with a frontal area of 30 ft2, traveling at 60 mph at sea level with a drag coefficient of 0.4, the drag would be about 110 lb. That would be about 17.7 horsepower or about 13.2 kW. B*f ffkAir Resistance 303Power needed to overcome air resistance increases with the cube of the vehicles velocity. Going from 60 to 75 mph is an air resistance power increase of 95% Energy to overcome air resistance to go a fixed distance, increases with the square of the vehicles velocity. (GPM or KW-hr)fjVery High DragWCd between 0.4 & 0.5?Y Cd about 0.19 Z Cd about 0.11  Climbing Hills03The maximum freeway grade is 6% Some San Diego roads have grades as high as 24%. The force needed for a 4000 lb vehicle to climb a 6% grade is 240 lb. To climb a 6% grade at 60 mph, A 4000 lb vehicle needs an additional 38.4 horsepower or about 29 kW more. f  Acceleration 03If you drop something, it will accelerate at the rate of about 22 mph/sec. This is know as a 1 g acceleration. An horizontal acceleration of half that, or about 10 mph/sec would be an  aggressive , typical of a sports car with a very fast driver. An acceleration of 2 mph/sec would be  conservative , typical of an older driver, or of a Honda Civic or VW GT. lZlf Acceleration 203An acceleration of 2.2 mph/sec, or 0.1 g, of a 4000 lb vehicle would require a force of 400 lb. At 60 mph, this would require an additional 64 horsepower or about 48 kW more. $f Losses03iRoughly 4.8 + 13.2 or 18.0 kW would be needed to maintain 60 mph on a level road with a 4000 lb vehicle with typical radial tires and a cross section of 30 ft2 with a Cd of 0.4. Roughly 18.0 + 28.6 or 46.6 kW would be needed to maintain 60 mph up a 6% grade. Roughly 47 + 36 or 83 kW would be needed to accelerate at 2.2 mph/sec up the 6% grade at 60 mph. FiZZf ff Losses 2 03Running 18 kW for 40 minutes run would be 12 kW-hr of energy for a distance of 40 miles at 60 mph. With a battery pack of 144 volts, this would be about 90 amp-hr of usage. For long life of a Lead-Acid battery, the depth of discharge should be less then 80%. Even an 80% DOD would shorten the life. A 100% DOD would give a very short life. Thus the need for at least a 120 amp-hr battery for the described vehicle. fZNB<Measure Losses03[It takes a force equal to the weight of the vehicle to cause a 1 g deceleration A 1 g deceleration is about 22 mph/sec Measure how long it takes on a level road to coast from 65 to 55 mph in sec (t) Deceleration (d) = (65-55)/t mph/sec Force (f) is vehicle weight * d/22 lbs Loss is about f * 60 /375 horsepower ( 1 hp = 375 lb-mph = 746 watts )&\[ff  Battery 1 03~The source of energy for an electric vehicles is its battery. The battery must supply enough current to the electric motor in order for it to supply the needed torque. The battery must have enough voltage to force the needed current through the electric motor for the desired speed. The battery must have enough energy to supply the needed power for the needed amount of time. (~ZZf  Battery 2 03"U. S. Battery makes an 8 volt battery with a 75 amp discharge time of 85 minutes called the US-8VGC. It has a weight of about 65 lb. 18 batteries in series will supply 144 volts. 18 batteries will weight about 1170 lb. Amp-Hr rating of about 106 min @ 75 A. (178 amp-hr @ 20 hr rate) #"f"8Battery 3  [rules of thumb]* 03$3Lead-acid batteries in an electric vehicle need to be at least 33% of a good vehicles gross weight to get a range of more than 40 miles with conservative driving. To get good performance, you need at least 33% of the vehicles gross weight to be active, on-line battery. $fJBattery 4  [lead-acid battery life]*& 03 3]Do not exceed 80% depth of discharge. Keep battery voltage within normal range. [For 144 V pack, keep pack above 120 V and below 185 V at all times.] Limit maximum current. [Excessive current leads to short life and even battery failure.] [Keep maximum current below the current that gives a full charge to 80% Discharge time of 20 minutes.] $]^f  Drive Train 03The electric motor must have enough torque to overcome the losses, climb hills and accelerate the vehicle to a useful speed. The electric motor must have enough speed for the vehicle. Gears are used to match the electric motor characteristics to the vehicle requirements. $f Drive Train 203Selected tire size is P185/60R14 Tire will make 888 revolutions per mile Each tire will hold 1047 lb at 35 psi Total gear ratio is 3.75:1 Motor RPM @ 60 mph is 3330 Maximum gross vehicle weight (including 143 lb motor, 1134 lb of batteries, 50 lb of controller & wiring, two 250 lb occupants and 250 lb of  stuff ) is 3700 lb. BKZZZZNfmElectric Motor03QSeries wound direct current motor In any gear, speed is proportional to RPM Constant torque for even acceleration Torque roughly proportional to current Increasing voltage is necessary to maintain current to maintain torque as vehicle speed and motor RPM increase Batteries must have enough voltage and current to maintain desired speed $QRfA;Electric Motor 203The selected electric motor is the Advanced DC FB1-4001 Diameter is 9.1 Weight is 143 lb Max continuous rated current is 180 A Max 1 hour rated current is 200 A Max 5 minute rated current is 340 A Current is limited by motor temperature Motor speed should be kept under 6000 rpm [High rpm causes rapid brush and bearing wear.]$MLf^Motor Characteristics(03{Torque increases with current. Back voltage increases with current and motor speed [rpm]. [Motors are also a generator].\| f$/$ff3fvVehicle Characteristics03 You select with your foot the current sent to the electric motor. With a constant current you have a constant torque. As the vehicle accelerates from a stop, the controller increases the voltage on the motor to maintain that current until there is no more voltage. [battery voltage reached] As the vehicle continues to accelerate, current and therefore torque decrease, causing acceleration to also decrease until torque is just enough to match losses and you maintain a constant speed. fwVehicle Characteristics 203k In the following graph, for a given foot setting, you follow a constant torque line up to the battery voltage and then follow a horizontal line to the right as rpm and vehicle speed increase. Note the corresponding decrease in torque. You must have enough battery voltage to push the current you need to get the torque you need to go the speed you need. (lfyjul Assumptions* 030Battery voltage is 144 volts. Maximum controller current is 500 amps. Motor is Advanced DC FB1. Vehicle gross weight is 4000 lb. Tire drag is 1% of vehicle weight. Aerodynamic Cd is 0.4. Cross sectional area is 30 ft2. Vehicle is at Sea Level.`ff ffC{FD>Warning0  Note that the highest force in the previous slide is for a current of almost 500 A that will quickly overheat the motor. The continuous current must be less then 180 A and that means that the continuous force must be less than 1/3 of the maximum force shown.   ~ Motor Comment03 Remember that power is the product of torque and rpm. With the ADC FB1-4001, the 200 A continuous rating is a torque limit of about 30 ft-lbs. At 30 ft-lbs, it takes about 144 V for a motor speed of 5500 rpm. This is about 31 hp. Actually, I2R losses in battery, controller and wiring will reduce the actual voltage available to the motor. At 80% DOD with a 200 A load, the maximum voltage at the motor may be as low as 120 V for only 4500 rpm. [25 hp]8f ffMotor Comment 203 Gearing the motor for 4500 rpm at the top vehicle speed [70 mph?] will take full advantage of the capability of the battery, controller and motor in the real world. Too many car conversions fail to take into account worst case conditions. [The last hill to climb with batteries at 80% DOD.] Of course some have the option to shift to a lower gear and struggle at a lower speed. ftMeasure Performance*030xIt takes a force equal to the weight of the vehicle in addition to the force to over-come losses to cause a 1 g acceleration. Measure how long it takes on a level road to accelerate from 55 to 65 mph in seconds (t). Acceleration (a) at 60 mph is about 10/t. Force (f) is about weight * a/22 lbs. Acceleration Hp is about f * 60 / 375. Total Hp is Acceleration Hp + Loss Hp. 6xZZwf)%Range*0304Now that we have a rough idea of the vehicle s performance, the next question is how far will it go on a charge. In other words, what is its range? Range should really be determined by how far it will go on 80% of a charge since completely discharging a battery will ruin it. Note that the capacity [amp-hr] decreases as the current increases. Also note that the voltage decreases as the charge is used up.(ffRange 2*030To estimate range at a given speed, determine the force needed at that speed. The force (lb) x speed (mph) / 375 is the hp needed to maintain that speed. Multiply hp by .746 to get kW. Divide kW by the battery voltage to get battery current. Estimate battery amp-hr at that current and divide by the current. Multiply hr by 0.8 to get the approximate number of hours. Multiply hours by the speed to get an estimate of range. 8fff+'Available Current*030\The total capacity of the battery is non-linear. The minutes the battery can provide power decreases faster then the amps supplied by the US 8V GC battery: 1041 minutes @ 10 amps 341 minutes @ 25 amps 146 minutes @ 50 amps 94 minutes @ 75 amps 66 minutes @ 100 amps 50 minutes @ 125 amps(]ffb83NOTICE03 The numbers used on the previous slides were taken from the best information and estimates available. Exact measured numbers were not available. Therefore, notice is given that the conclusions are approximate ballpark estimates. Actual performance to be determined."  f94Charging 0* U.S. Battery recommends that: Voltage not exceed 2.585 V per cell Current not exceed AH/10 Time not exceed 10 hours http://www.usbattery.com/pages/usbspecs.htm In other words, for a  144 volt pack, the charging current should not exceed 165/10 or about 16.5 amps until limited by the total voltage that must not exceed 186 volts. Maximum charge time is 10 hours. Check water level after charge.  %$fY+f>  :5Charging 0m186 volts times 16.5 amps is 3065 watts. That would be about 26 amps from a 120 volt source, or 13 amps from a 240 volts source, not taking into account efficiency of the charger. If time were short, the batteries could be charged at 25 amps. That would be 4650 watts, or almost 20 amps from a 240 volt source. A 30 amp 240 volt service is best for charging. \m$jfk<7. . . ie Hybrid0CFor long trips a small motor-generator can be added to extend range. Motor generators are made to run on a variety of different fuels. Commercial motor-generators include gasoline, diesel, propane, etc. Be sure the controller can take the higher voltage. Voltage should not exceed the maximum battery charging voltage. >D$B(|K fuSan Diego Car Conversion Project July-August 2008 Physics of Electric Vehicles by Russ Lemon Russ@FarTooMuch.Info fv4f(J J f ff"` 0`taThe End`  To Return http://EVAoSD.com $ 0/dKLMNOPQ R S T VWYZ[\h j"k#o'q)r*t,z014;@ABCA p`0(  `x ` c $0`Ppp 0 x ` c $0P 0 H ` 0޽h ? ff3333f1 0 0(   x   c $X0 0 x   c $0 0 H   0޽h ? ff3333f  z`X$(  Xr X S C#Pp  # r X S hD#@0  # 6 X S  ? ff333fr |#+Dw( 5  6"http://EVAoSD.com< (Russ@FarTooMuch.Info<(Russ@FarTooMuch.Infol/ 0DTimes New Roman`F|d0|Wo 0DComic Sans MSn`F|d0|Wo 0Bg33(.  @n?" dd@  @@`` |V  *,67I 9-Q!"?!#%'),.2579@F?AGR 2T> 2:= %)D#" ABV GH/01xb$ItS5T[Y $ $ b$/71q6aTIY $ $ $ b$i ƌE7 Έ)_ $ $ $ $ R$2KwsR $ R$[ JY囲yĠ(H1 $ $ $ $ $ $ $ R$+GsmjKOlz $ $ R$0$~r>\k}s  0e0e     A@f A5% 8c8c     ?1 d0u0@Ty2 NP'p<'pA)BCD|E||S" J /@3q$ ʚ;^ :ʚ;g4dddd0pppp@ <4!d!d 0,LG<4dddd 0,LG 2*___PPT9 z? %``The Physics ofHElectric VehiclesH}Circle Calibrate your video screen or projector. The next slide must show as a circle for the pictures to have the correct aspect ratio.|zThe Physics ofH6Electric Vehicles By Russ Lemon Russ@FarTooMuch.Info Z7H(  "!  0!5iHow they work & And why&fxElectric Vehicles0yWhat follows is a discussion of how chemical energy is converted into electric energy and then into mechanical energy to propel a vehicle. The discussion includes how mechanical energy is used to overcome the vehicle losses of tire and aerodynamic drag, and yet have enough energy left over to climb hills and accelerate the vehicle. To go fast and far, minimize your losses.&zPf*yWilliam Thomson$ [aka Lord Kelvin] 1824-1907  When you can measure what you are speaking about, and express it in numbers, you know something about it; but when you cannot measure it, when you cannot express it in numbers, your knowledge is of a meager and unsatisfactory kind: it may be the beginning of knowledge, but you have scarcely, in your thoughts, advanced to the stage of science. V+^f{ Basic Units 3MA decimeter is a tenth of a meter, or about 3 & 15/16 inches A cubic decimeter is a liter A liter of cold water has about 1 kg of mass In San Diego that 1 kg of mass has a weight of about 9.8 newtons (force) Raise it up 1 meter and you have done 9.8 joules of work. Raise it up 1 meter in one second requires a power of 9.8 watts. VNZ0},ICEnergy3energy = force x distance joules = newtons x meters joules = volts x coulombs 1 kW-hr = 3.6 MJ [megajoules] 1 hp-hr about 2.7 MJ 1 BTU about 1054.8 J $,#7 .UPower3power = force x speed watts = newtons x meters/second watts = volts x amps one horsepower = 746 watts one horsepower = lbf x mph / 375 (i.e. 1 hp = 5 lbf @ 75 mph) 1 hp = ft-lb x rpm / 5252 >R&Rolling Resistance03`Rolling resistance of an average Radial Ply Passenger Tire inflated to 32 psi is about 1% of the weight on the tire. Rolling resistance of an average Bias Ply Tire can be more than double that of a radial ply tire with the same load and pressure. Rolling resistance is measured at maximum inflation pressure and increases as tire pressure decreases. aZaf,JRolling Resistance 203For a vehicle weighing 4000 lb, a rolling resistance of 1% of load represents a drag of 40 lb. At 60 mph, a drag of 40 lb represents a loss of 6.4 horsepower or about 4.8 kW. There are now special low rolling resistance passenger tires with a rolling resistance as low as 0.6% of load. fAir Resistance03CAir Resistance is proportional to the density of the air, the drag coefficient of the vehicle, the frontal area of the vehicle, and the speed of the vehicle squared. Typical Coefficient of Drag (Cd) for a modern passenger vehicle [with windows rolled up] is about 0.4. The EV1 was about .19. The Aptera is about .11 DDf, Air Resistance 203For a vehicle with a frontal area of 30 ft2, traveling at 60 mph at sea level with a drag coefficient of 0.4, the drag would be about 110 lb. That would be about 17.7 horsepower or about 13.2 kW. B*f ffkAir Resistance 303Power needed to overcome air resistance increases with the cube of the vehicles velocity. Going from 60 to 75 mph is an air resistance power increase of 95% Energy to overcome air resistance to go a fixed distance, increases with the square of the vehicles velocity. (GPM or KW-hr)fjVery High DragWCd between 0.4 & 0.5?Y Cd about 0.19 Z Cd about 0.11  Climbing Hills03The maximum freeway grade is 6% Some San Diego roads have grades as high as 24%. The force needed for a 4000 lb vehicle to climb a 6% grade is 240 lb. To climb a 6% grade at 60 mph, A 4000 lb vehicle needs an additional 38.4 horsepower or about 29 kW more. f  Acceleration 03If you drop something, it will accelerate at the rate of about 22 mph/sec. This is know as a 1 g acceleration. An horizontal acceleration of half that, or about 10 mph/sec would be an  aggressive , typical of a sports car with a very fast driver. An acceleration of 2 mph/sec would be  conservative , typical of an older driver, or of a Honda Civic or VW GT. lZlf Acceleration 203An acceleration of 2.2 mph/sec, or 0.1 g, of a 4000 lb vehicle would require a force of 400 lb. At 60 mph, this would require an additional 64 horsepower or about 48 kW more. $f Losses03iRoughly 4.8 + 13.2 or 18.0 kW would be needed to maintain 60 mph on a level road with a 4000 lb vehicle with typical radial tires and a cross section of 30 ft2 with a Cd of 0.4. Roughly 18.0 + 28.6 or 46.6 kW would be needed to maintain 60 mph up a 6% grade. Roughly 47 + 36 or 83 kW would be needed to accelerate at 2.2 mph/sec up the 6% grade at 60 mph. FiZZf ff Losses 2 03Running 18 kW for 40 minutes run would be 12 kW-hr of energy for a distance of 40 miles at 60 mph. With a battery pack of 144 volts, this would be about 90 amp-hr of usage. For long life of a Lead-Acid battery, the depth of discharge should be less then 80%. Even an 80% DOD would shorten the life. A 100% DOD would give a very short life. Thus the need for at least a 120 amp-hr battery for the described vehicle. fZNB<Measure Losses03[It takes a force equal to the weight of the vehicle to cause a 1 g deceleration A 1 g deceleration is about 22 mph/sec Measure how long it takes on a level road to coast from 65 to 55 mph in sec (t) Deceleration (d) = (65-55)/t mph/sec Force (f) is vehicle weight * d/22 lbs Loss is about f * 60 /375 horsepower ( 1 hp = 375 lb-mph = 746 watts )&\[ff  Battery 1 03~The source of energy for an electric vehicles is its battery. The battery must supply enough current to the electric motor in order for it to supply the needed torque. The battery must have enough voltage to force the needed current through the electric motor for the desired speed. The battery must have enough energy to supply the needed power for the needed amount of time. (~ZZf  Battery 2 03"U. S. Battery makes an 8 volt battery with a 75 amp discharge time of 85 minutes called the US-8VGC. It has a weight of about 65 lb. 18 batteries in series will supply 144 volts. 18 batteries will weight about 1170 lb. Amp-Hr rating of about 106 min @ 75 A. (178 amp-hr @ 20 hr rate) #"f"8Battery 3  [rules of thumb]* 03$3Lead-acid batteries in an electric vehicle need to be at least 33% of a good vehicles gross weight to get a range of more than 40 miles with conservative driving. To get good performance, you need at least 33% of the vehicles gross weight to be active, on-line battery. $fJBattery 4  [lead-acid battery life]*& 03 3]Do not exceed 80% depth of discharge. Keep battery voltage within normal range. [For 144 V pack, keep pack above 120 V and below 185 V at all times.] Limit maximum current. [Excessive current leads to short life and even battery failure.] [Keep maximum current below the current that gives a full charge to 80% Discharge time of 20 minutes.] $]^f  Drive Train 03The electric motor must have enough torque to overcome the losses, climb hills and accelerate the vehicle to a useful speed. The electric motor must have enough speed for the vehicle. Gears are used to match the electric motor characteristics to the vehicle requirements. $f Drive Train 203Selected tire size is P185/60R14 Tire will make 888 revolutions per mile Each tire will hold 1047 lb at 35 psi Total gear ratio is 3.75:1 Motor RPM @ 60 mph is 3330 Maximum gross vehicle weight (including 143 lb motor, 1134 lb of batteries, 50 lb of controller & wiring, two 250 lb occupants and 250 lb of  stuff ) is 3700 lb. BKZZZZNfmElectric Motor03QSeries wound direct current motor In any gear, speed is proportional to RPM Constant torque for even acceleration Torque roughly proportional to current Increasing voltage is necessary to maintain current to maintain torque as vehicle speed and motor RPM increase Batteries must have enough voltage and current to maintain desired speed $QRfA;Electric Motor 203The selected electric motor is the Advanced DC FB1-4001 Diameter is 9.1 Weight is 143 lb Max continuous rated current is 180 A Max 1 hour rated current is 200 A Max 5 minute rated current is 340 A Current is limited by motor temperature Motor speed should be kept under 6000 rpm [High rpm causes rapid brush and bearing wear.]$MLf^Motor Characteristics(03{Torque increases with current. Back voltage increases with current and motor speed [rpm]. [Motors are also a generator].\| f$/$ff3fvVehicle Characteristics03 You select with your foot the current sent to the electric motor. With a constant current you have a constant torque. As the vehicle accelerates from a stop, the controller increases the voltage on the motor to maintain that current until there is no more voltage. [battery voltage reached] As the vehicle continues to accelerate, current and therefore torque decrease, causing acceleration to also decrease until torque is just enough to match losses and you maintain a constant speed. fwVehicle Characteristics 203k In the following graph, for a given foot setting, you follow a constant torque line up to the battery voltage and then follow a horizontal line to the right as rpm and vehicle speed increase. Note the corresponding decrease in torque. You must have enough battery voltage to push the current you need to get the torque you need to go the speed you need. (lfyjul Assumptions* 030Battery voltage is 144 volts. Maximum controller current is 500 amps. Motor is Advanced DC FB1. Vehicle gross weight is 4000 lb. Tire drag is 1% of vehicle weight. Aerodynamic Cd is 0.4. Cross sectional area is 30 ft2. Vehicle is at Sea Level.`ff ffC{FD>Warning0  Note that the highest force in the previous slide is for a current of almost 500 A that will quickly overheat the motor. The continuous current must be less then 180 A and that means that the continuous force must be less than 1/3 of the maximum force shown.   ~ Motor Comment03 Remember that power is the product of torque and rpm. With the ADC FB1-4001, the 200 A continuous rating is a torque limit of about 30 ft-lbs. At 30 ft-lbs, it takes about 144 V for a motor speed of 5500 rpm. This is about 31 hp. Actually, I2R losses in battery, controller and wiring will reduce the actual voltage available to the motor. At 80% DOD with a 200 A load, the maximum voltage at the motor may be as low as 120 V for only 4500 rpm. [25 hp]8f ffMotor Comment 203 Gearing the motor for 4500 rpm at the top vehicle speed [70 mph?] will take full advantage of the capability of the battery, controller and motor in the real world. Too many car conversions fail to take into account worst case conditions. [The last hill to climb with batteries at 80% DOD.] Of course some have the option to shift to a lower gear and struggle at a lower speed. ftMeasure Performance*030xIt takes a force equal to the weight of the vehicle in addition to the force to over-come losses to cause a 1 g acceleration. Measure how long it takes on a level road to accelerate from 55 to 65 mph in seconds (t). Acceleration (a) at 60 mph is about 10/t. Force (f) is about weight * a/22 lbs. Acceleration Hp is about f * 60 / 375. Total Hp is Acceleration Hp + Loss Hp. 6xZZwf)%Range*0304Now that we have a rough idea of the vehicle s performance, the next question is how far will it go on a charge. In other words, what is its range? Range should really be determined by how far it will go on 80% of a charge since completely discharging a battery will ruin it. Note that the capacity [amp-hr] decreases as the current increases. Also note that the voltage decreases as the charge is used up.(ffRange 2*030To estimate range at a given speed, determine the force needed at that speed. The force (lb) x speed (mph) / 375 is the hp needed to maintain that speed. Multiply hp by .746 to get kW. Divide kW by the battery voltage to get battery current. Estimate battery amp-hr at that current and divide by the current. Multiply hr by 0.8 to get the approximate number of hours. Multiply hours by the speed to get an estimate of range. 8fff+'Available Current*030\The total capacity of the battery is non-linear. The minutes the battery can provide power decreases faster then the amps supplied by the US 8V GC battery: 1041 minutes @ 10 amps 341 minutes @ 25 amps 146 minutes @ 50 amps 94 minutes @ 75 amps 66 minutes @ 100 amps 50 minutes @ 125 amps(]ffb83NOTICE03 The numbers used on the previous slides were taken from the best information and estimates available. Exact measured numbers were not available. Therefore, notice is given that the conclusions are approximate ballpark estimates. Actual performance to be determined."  f94Charging 0, U.S. Battery recommends that: Voltage not exceed 2.585 V per cell Current not exceed AH/10 Time not exceed 10 hours http://www.usbattery.com/pages/usbspecs.htm In other words, for a  144 volt pack, the charging current should not exceed 165/10 or about 16.5 amps until limited by the total voltage that must not exceed 186 volts. Maximum charge time is 10 hours. Check water level after charge.  %$fY+f>  :5Charging 0m186 volts times 16.5 amps is 3065 watts. That would be about 26 amps from a 120 volt source, or 13 amps from a 240 volts source, not taking into account efficiency of the charger. If time were short, the batteries could be charged at 25 amps. That would be 4650 watts, or almost 20 amps from a 240 volt source. A 30 amp 240 volt service is best for charging. \m$jfk<7. . . ie Hybrid0CFor long trips a small motor-generator can be added to extend range. Motor generators are made to run on a variety of different fuels. Commercial motor-generators include gasoline, diesel, propane, etc. Be sure the controller can take the higher voltage. Voltage should not exceed the maximum battery charging voltage. >D$B(|K fuSan Diego Car Conversion Project July-August 2008 Physics of Electric Vehicles by Russ Lemon Russ@FarTooMuch.Info fv4f(J J f ff"` 0`taThe End`  To Return http://EVAoSD.com $ 0/dKLMNOPQ R S T VWYZ[\h j"k#o'q)r*t,z014;@ABC) 0(  x  c $ #`Pp # x  c $< # # H  0޽h ? ff3333frA9 `Dv( 5  6"http://EVAoSD.com< (Russ@FarTooMuch.Info<(Russ@FarTooMuch.Infol/ 0DTimes New Roman<\|d0|Wo 0DComic Sans MSn<\|d0|Wo 0Bg33(.  @n?" dd@  @@`` |V   )+6:I <,Q !? "$&(+-3579@FBDGR  5T> 5 :@ $(D"! AB1GH..0/xb$ItS5T[Y$$b$/71q6aTIY$$$b$i ƌE7 Έ)_$$$$R$2KwsR$R$[ JY囲yĠ(H1$$$$$$$R$+GsmjKOlz$$R$0$~r>\k}s 0e0e     A@f A5% 8c8c     ?1 d0u0@Ty2 NP'p<'pA)BCD|E||S" J /@3q$ ʚ;^ :ʚ;g4dddd0p ppp@ <4!d!d 0,(]<4dddd 0,(] 2*___PPT9 z? %#``The Physics ofHElectric VehiclesH}Circle Calibrate your video screen or projector. The next slide must show as a circle for the pictures to have the correct aspect ratio.|zThe Physics ofH6Electric Vehicles By Russ Lemon Russ@FarTooMuch.Info Z7H(  "!  0!5iHow they work & And why&fxElectric Vehicles0yWhat follows is a discussion of how chemical energy is converted into electric energy and then into mechanical energy to propel a vehicle. The discussion includes how mechanical energy is used to overcome the vehicle losses of tire and aerodynamic drag, and yet have enough energy left over to climb hills and accelerate the vehicle. To go fast and far, minimize your losses.&zPf*yWilliam Thomson$ [aka Lord Kelvin] 1824-1907  When you can measure what you are speaking about, and express it in numbers, you know something about it; but when you cannot measure it, when you cannot express it in numbers, your knowledge is of a meager and unsatisfactory kind: it may be the beginning of knowledge, but you have scarcely, in your thoughts, advanced to the stage of science. V+^f{ Basic Units 32A decimeter is a tenth of a meter A cubic decimeter is a liter A liter of cold water has about 1 kg of mass In San Diego that 1 kg of mass has a weight of about 9.8 newtons (force) Raise it up 1 meter and you have done 9.8 joules of work. Raise it up 1 meter in one second requires a power of 9.8 watts. 63}1ICEnergy3energy = force x distance joules = newtons x meters joules = volts x coulombs 1 kW-hr = 3.6 MJ [megajoules] 1 BTU about 1054.8 J $a UPower3fpower = force x speed watts = newtons x meters/second watts = volts x amps one horsepower = 746 watts $fgRolling Resistance03aRolling resistance of an average Radial Ply Passenger Tire inflated to 220 kPa is about 1% of the weight on the tire. Rolling resistance of an average Bias Ply Tire can be more than double that of a radial ply tire with the same load and pressure. Rolling resistance is measured at maximum inflation pressure and increases as tire pressure decreases. bZbf,KRolling Resistance 203-For a 2000 kg vehicle, a weight of almost 20 kN, a rolling resistance of 1% of load represents a drag of almost 200 N. At 100 kM/Hr, a drag of almost 200 N represents a loss of about 5.4 kW. There are now special low rolling resistance passenger tires with a rolling resistance as low as 0.6% of load. .-f,-OAir Resistance03EAir Resistance is proportional to the density of the air, the drag coefficient of the vehicle, the frontal area of the vehicle, and the speed of the vehicle squared. Typical Coefficient of Drag (Cd) for a modern passenger vehicle [with windows rolled up] is about 0.4. The EV1 was about .19. The Aptera 2 is about .15 FFf, Air Resistance 203For a vehicle with a frontal area of 30 m2, traveling at 100 km/Hr at sea level with a drag coefficient of 0.4, the drag would be about xxx N. That would be about about xx13.2 kW. B)f ff,)kAir Resistance 303Power needed to overcome air resistance increases with the cube of the vehicles velocity. Going from 60 to 75 mph is an air resistance power increase of 95% Energy to overcome air resistance to go a fixed distance, increases with the square of the vehicles velocity. (GPM or KW-hr)fjVery High DragWCd between 0.4 & 0.5?Y Cd about 0.19 Z Cd about 0.11  Climbing Hills03The maximum freeway grade is 6% Some San Diego roads have grades as high as 24%. The force needed for a 4000 lb vehicle to climb a 6% grade is 240 lb. To climb a 6% grade at 60 mph, A 4000 lb vehicle needs an additional 38.4 horsepower or about 29 kW more. f  Acceleration 03If you drop something, it will accelerate at the rate of about 22 mph/sec. This is know as a 1 g acceleration. An horizontal acceleration of half that, or about 10 mph/sec would be an  aggressive , typical of a sports car with a very fast driver. An acceleration of 2 mph/sec would be  conservative , typical of an older driver, or of a Honda Civic or VW GT. lZlf Acceleration 203An acceleration of 2.2 mph/sec, or 0.1 g, of a 4000 lb vehicle would require a force of 400 lb. At 60 mph, this would require an additional 64 horsepower or about 48 kW more. $f Losses03iRoughly 4.8 + 13.2 or 18.0 kW would be needed to maintain 60 mph on a level road with a 4000 lb vehicle with typical radial tires and a cross section of 30 ft2 with a Cd of 0.4. Roughly 18.0 + 28.6 or 46.6 kW would be needed to maintain 60 mph up a 6% grade. Roughly 47 + 36 or 83 kW would be needed to accelerate at 2.2 mph/sec up the 6% grade at 60 mph. FiZZf ff Losses 2 03Running 18 kW for 40 minutes run would be 12 kW-hr of energy for a distance of 40 miles at 60 mph. With a battery pack of 144 volts, this would be about 90 amp-hr of usage. For long life of a Lead-Acid battery, the depth of discharge should be less then 80%. Even an 80% DOD would shorten the life. A 100% DOD would give a very short life. Thus the need for at least a 120 amp-hr battery for the described vehicle. fZNB<Measure Losses03[It takes a force equal to the weight of the vehicle to cause a 1 g deceleration A 1 g deceleration is about 22 mph/sec Measure how long it takes on a level road to coast from 65 to 55 mph in sec (t) Deceleration (d) = (65-55)/t mph/sec Force (f) is vehicle weight * d/22 lbs Loss is about f * 60 /375 horsepower ( 1 hp = 375 lb-mph = 746 watts )&\[ff  Battery 1 03~The source of energy for an electric vehicles is its battery. The battery must supply enough current to the electric motor in order for it to supply the needed torque. The battery must have enough voltage to force the needed current through the electric motor for the desired speed. The battery must have enough energy to supply the needed power for the needed amount of time. (~ZZf  Battery 2 03"U. S. Battery makes an 8 volt battery with a 75 amp discharge time of 85 minutes called the US-8VGC. It has a weight of about 65 lb. 18 batteries in series will supply 144 volts. 18 batteries will weight about 1170 lb. Amp-Hr rating of about 106 min @ 75 A. (178 amp-hr @ 20 hr rate) #"f"8Battery 3  [rules of thumb]* 03$3Lead-acid batteries in an electric vehicle need to be at least 33% of a good vehicles gross weight to get a range of more than 40 miles with conservative driving. To get good performance, you need at least 33% of the vehicles gross weight to be active, on-line battery. $fJBattery 4  [lead-acid battery life]*& 03 3]Do not exceed 80% depth of discharge. Keep battery voltage within normal range. [For 144 V pack, keep pack above 120 V and below 185 V at all times.] Limit maximum current. [Excessive current leads to short life and even battery failure.] [Keep maximum current below the current that gives a full charge to 80% Discharge time of 20 minutes.] $]^f  Drive Train 03The electric motor must have enough torque to overcome the losses, climb hills and accelerate the vehicle to a useful speed. The electric motor must have enough speed for the vehicle. Gears are used to match the electric motor characteristics to the vehicle requirements. $f Drive Train 203Selected tire size is P185/60R14 Tire will make 888 revolutions per mile Each tire will hold 1047 lb at 35 psi Total gear ratio is 3.75:1 Motor RPM @ 60 mph is 3330 Maximum gross vehicle weight (including 143 lb motor, 1134 lb of batteries, 50 lb of controller & wiring, two 250 lb occupants and 250 lb of  stuff ) is 3700 lb. BKZZZZNfmElectric Motor03QSeries wound direct current motor In any gear, speed is proportional to RPM Constant torque for even acceleration Torque roughly proportional to current Increasing voltage is necessary to maintain current to maintain torque as vehicle speed and motor RPM increase Batteries must have enough voltage and current to maintain desired speed $QRfA;Electric Motor 203The selected electric motor is the Advanced DC FB1-4001 Diameter is 9.1 Weight is 143 lb Max continuous rated current is 180 A Max 1 hour rated current is 200 A Max 5 minute rated current is 340 A Current is limited by motor temperature Motor speed should be kept under 6000 rpm [High rpm causes rapid brush and bearing wear.]$MLf^Motor Characteristics(03{Torque increases with current. Back voltage increases with current and motor speed [rpm]. [Motors are also a generator].\| f$/$ff3fvVehicle Characteristics03 You select with your foot the current sent to the electric motor. With a constant current you have a constant torque. As the vehicle accelerates from a stop, the controller increases the voltage on the motor to maintain that current until there is no more voltage. [battery voltage reached] As the vehicle continues to accelerate, current and therefore torque decrease, causing acceleration to also decrease until torque is just enough to match losses and you maintain a constant speed. fwVehicle Characteristics 203k In the following graph, for a given foot setting, you follow a constant torque line up to the battery voltage and then follow a horizontal line to the right as rpm and vehicle speed increase. Note the corresponding decrease in torque. You must have enough battery voltage to push the current you need to get the torque you need to go the speed you need. (lfyjul Assumptions* 030Battery voltage is 144 volts. Maximum controller current is 500 amps. Motor is Advanced DC FB1. Vehicle gross weight is 4000 lb. Tire drag is 1% of vehicle weight. Aerodynamic Cd is 0.4. Cross sectional area is 30 ft2. Vehicle is at Sea Level.`ff ffC{FD>Warning0  Note that the highest force in the previous slide is for a current of almost 500 A that will quickly overheat the motor. The continuous current must be less then 180 A and that means that the continuous force must be less than 1/3 of the maximum force shown.   ~ Motor Comment03 Remember that power is the product of torque and rpm. With the ADC FB1-4001, the 200 A continuous rating is a torque limit of about 30 ft-lbs. At 30 ft-lbs, it takes about 144 V for a motor speed of 5500 rpm. This is about 31 hp. Actually, I2R losses in battery, controller and wiring will reduce the actual voltage available to the motor. At 80% DOD with a 200 A load, the maximum voltage at the motor may be as low as 120 V for only 4500 rpm. [25 hp]8f ffMotor Comment 203 Gearing the motor for 4500 rpm at the top vehicle speed [70 mph?] will take full advantage of the capability of the battery, controller and motor in the real world. Too many car conversions fail to take into account worst case conditions. [The last hill to climb with batteries at 80% DOD.] Of course some have the option to shift to a lower gear and struggle at a lower speed. ftMeasure Performance*030xIt takes a force equal to the weight of the vehicle in addition to the force to over-come losses to cause a 1 g acceleration. Measure how long it takes on a level road to accelerate from 55 to 65 mph in seconds (t). Acceleration (a) at 60 mph is about 10/t. Force (f) is about weight * a/22 lbs. Acceleration Hp is about f * 60 / 375. Total Hp is Acceleration Hp + Loss Hp. 6xZZwf)%Range*0304Now that we have a rough idea of the vehicle s performance, the next question is how far will it go on a charge. In other words, what is its range? Range should really be determined by how far it will go on 80% of a charge since completely discharging a battery will ruin it. Note that the capacity [amp-hr] decreases as the current increases. Also note that the voltage decreases as the charge is used up.(ffRange 2*030To estimate range at a given speed, determine the force needed at that speed. The force (lb) x speed (mph) / 375 is the hp needed to maintain that speed. Multiply hp by .746 to get kW. Divide kW by the battery voltage to get battery current. Estimate battery amp-hr at that current and divide by the current. Multiply hr by 0.8 to get the approximate number of hours. Multiply hours by the speed to get an estimate of range. 8fff+'Available Current*030\The total capacity of the battery is non-linear. The minutes the battery can provide power decreases faster then the amps supplied by the US 8V GC battery: 1041 minutes @ 10 amps 341 minutes @ 25 amps 146 minutes @ 50 amps 94 minutes @ 75 amps 66 minutes @ 100 amps 50 minutes @ 125 amps(]ffb83NOTICE03 The numbers used on the previous slides were taken from the best information and estimates available. Exact measured numbers were not available. Therefore, notice is given that the conclusions are approximate ballpark estimates. Actual performance to be determined."  f94Charging 0, U.S. Battery recommends that: Voltage not exceed 2.585 V per cell Current not exceed AH/10 Time not exceed 10 hours http://www.usbattery.com/pages/usbspecs.htm In other words, for a  144 volt pack, the charging current should not exceed 165/10 or about 16.5 amps until limited by the total voltage that must not exceed 186 volts. Maximum charge time is 10 hours. Check water level after charge.  %$ f Y +f>  :5Charging 0m186 volts times 16.5 amps is 3065 watts. That would be about 26 amps from a 120 volt source, or 13 amps from a 240 volts source, not taking into account efficiency of the charger. If time were short, the batteries could be charged at 25 amps. That would be 4650 watts, or almost 20 amps from a 240 volt source. A 30 amp 240 volt service is best for charging. \m$jfk<7. . . ie Hybrid0CFor long trips a small motor-generator can be added to extend range. Motor generators are made to run on a variety of different fuels. Commercial motor-generators include gasoline, diesel, propane, etc. Be sure the controller can take the higher voltage. Voltage should not exceed the maximum battery charging voltage. >D