Basic Motion Calculations

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Basic Motion calculations.

Forces needed to move the Vehicle

Prepared for the Florida EAA

October 2008

David Kerzel



Basic Information about the Vehicle

� Weight of the car

� Maximum load it can carry

� Tire diameter � Final drive ratio

� Gear ratio for each transmission speed.



Linear Inertia

Inertia needs to be over come to accelerate

It all begins with Newton¶s Second Law

F=Ma

F in Lbf, M in Lb, a in ft/sec2

Most people think acceleration in mph/sec

To convert MPH/sec

divide MPH/sec by 21.95 => ft/sec2

M Mass is the weight of the vehicle in Lb after

conversion with batteries and occupants.



Linear Inertia

For a good starting point use the curb weight and

the maximum load weight, mine is 3,800 Lb

F=Ma

My driving is at 35-45 MPH and it takes about 10

seconds for me to get to 40 MPH.

That leads to an acceleration of 4 MPH/sec.

Acceleration needs to be ft/sec2 so divide 4MPH/sec by 21.95 to get 0.1822 ft/sec2.

Fal=Ma = 3800 X .1822 = 692Lbf



Rotating Inertia

All rotating parts also need to be accelerated. The wheelsand ties are obvious, but the axels and shafts. Motor armature and fly wheel all must be considered. Thesecalculations are impossible without weights anddimensions of the items. This rotational inertia is typical

5 to 20% of the linear inertia for production vehicles.Whether you are moving the car with the motor or the car is

causing the motor to rotate this rotational inertia is thesame so we can look at this like a force and make anassumption it will be 10%.

Far = 0.10 X Fal = 69Lbf

Far = 0.10 X 692 = 69Lbf



Aerodynamic Drag

As the car moves through the air there is moreforce needed.

Fdrag = (Cd X A X v2)/391

Cd is the drag coefficient A is frontal area of the

car in ft2 v is speed in MPH 391 fixes the

units

For my donor car it is Cd = 0.34 and A is 20 ft2

Speed 10 20 30 40 50 60 70 80

Fdrag 2 7 16 28 43 63 85 111



Wind Drag

Wind causes more aerodynamic drag.

Basically a head wind is like moving faster.

So driving at 40M

PH with a 10M

PH headwind give the aerodynamic drag of 50

MPH. similarly a tail wind reduces

aerodynamic drag. I am going to ignore it

because it is a small factor in all this.



Inclines

Even here in South Florida there are some minor inclines and a few steeper ones. Some Railroad crossings are steep, entrance ramps for I-95, where I work has a 25% grade from street to

parking lot.Fclimb = W X sin ()

= ArcTan (rise/run)

For 10% we get a angle of 8Û32¶

The sin of 8Û32¶ is 0.09963800 X 0.096 = 378Lbf

I would estimate 5% would be reasonable



Inclines

A 5% incline needs more force than theaerodynamic drag at 80 MPH

Grade sin () Weight Fdrag5% 0.050 3800 190

10% 0.100 3800 378

20% 0.196 3800 746

30% 0.287 3800 1092

40% 0.371 3800 1411



Rolling Drag

Tires have rolling resistance that ends up being aform of drag. Different tire designs have higher or lower rolling resistance. Higher air pressurereduces rolling resistance, there are tradeoffs.

Frolling = Cr X W X cos ()The angle in the formula reduces the rolling

resistance as the road incline angle increases.

Low rolling resistance tires have Cr of about 0.01

and normal tires are 0.02.On a flat surface:

Frolling = 0.02 X 3800 X 1 = 76Lbf



Total Acceleration Force

All of the above forces need to be added up for the totalforce.

These are the 4 MPH/sec acceleration forces. As long asforce can be provided the acceleration will continue. No

power source has unlimited force and at some point the

required force cannot be provided.

Speed 0 10 20 30 40 50 60 70 80

Fal 692 692 692 692 692 692 692 692 692

Far 69 69 69 69 69 69 69 69 69

Fdrag 0 2 9 20 35 54 78 107 139

Fclimb 190 190 190 190 190 190 190 190 190

Frolling 76 76 76 76 76 76 76 76 76

Ftotal 1027 1029 1036 1047 1062 1081 1105 1134 1166



Constant Velocity Force

Once cruising speed is reached the accelerationforces no longer apply so the totals reduce.

I have also eliminated the climbing factor.

These cruising forces show how traveling at 70

MPH requires about two times the force at 20MPH.

It shows how acceleration requires 5 to 10 times the

force that cursing at constant speed does.

Speed 10 20 30 40 50 60 70 80

Fdrag 2 9 20 35 54 78 107 139

Frolling 76 76 76 76 76 76 76 76

Ftotal 78 85 96 111 130 154 183 215



Torque at the wheels

Regardless of the number of driven wheels the sameamount of torque needs to be delivered to thewheels.

Torque is measured in Ft-Lb which is a weight of I Lb at

the end of a 1 foot lever.In a vehicle on wheels the length of the lever arm is the

radius, half the diameter of the tire, and the weight isthe required force.

My tires have a radius or 12.45 inches or 1.0375 ft.

If I combine that with the 1027Lbf from the accelerationcalculations I get

1027 X 1.0375 = 1065 ft-Lb of torque for starting andaccelerating at 4MPH/sec.



Torque at the motor Acceleration

The gears in the final drive and transmission tradetorque for shaft speeds. As the speed isdecreased the torque increases at the sameratio.

I need 1065 Ff-Lb at the wheels for 4 MPH/secacceleration. My final drive is 3.87:1 ratio sothe input torque is 1065 / 3.87 = 275 Ft-Lb to thefinal drive.

There is still the transmission between the motor and the final drive.

The final drive and transmission are not 100%efficient but this will be ignored for simplicity.




There is still the transmission between the motor

and the final drive

Output Torque Gear Ratio Input Torque

275 1 3.58 77

275 2 1.95 141

275 3 1.38 200

275 4 1.03 267

275 5 0.77 358




A 9 inch Advance DC motor has 240 Ft-Lb peaktorque at low speed. It would give the initial

acceleration well in 3nd gear. In 2nd gear more

than target acceleration could be reached and

maintained as speed increases.Output Torque Gear Ratio Input Torque

275 1 3.58 77

275 2 1.95 141

275 3 1.38 200

275 4 1.03 267

275 5 0.77 358

2nd gear will actually provide 6.8 MPH/Sec Acceleration to 10 MPH



Motor Torque Curve




The AC motor and controller I plan to use at 150% continuous torquehas 90 Ft-Lb output so I will need to start in first gear. The AC motor does not have the incredibly high low speed torque that a serieswound DC motor has. My motor is capable of higher torque butgetting a controller is the issue today. The AC advantage is thetorque stays constant over the full rated speed range. A Series

wound DC motor starts with very high torque but that torque isreduced as RPM increases.


275 1 3.58 77

275 2 1.95 141275 3 1.38 200

275 4 1.03 267

275 5 0.77 358



Power & Acceleration

Power is Torque X RPM / 5252

Power in in HP, Torque is in Ft-Lb

If the acceleration was reduced to 2MPH/sec the Fal andFar inertia forces would be half of the original levels.Everything else would be the same.

It would take twice as long to get to speed.

With a electric vehicle it takes the same amount of power toget to a speed. There is no penalty for aggressive

acceleration in electric vehicles. Actually rapid acceleration may be slightly more efficient

since less time is spent in acceleration.



Power at Constant Speed

More speed still requires more power.

Speed 10 20 30 40 50 60 70 80

Fdrag 2 9 20 35 54 78 107 139

Frolling 76 76 76 76 76 76 76 76Ftotal 78 85 96 111 130 154 183 215

Wheel RPM 135 270 405 540 675 810 945 1081

Power (HP) 2 5 8 12 17 25 34 46

Power (KW) 2 3 6 9 13 18 25 34



Vehicle Speed

Vehicle speed was in a few calculations but has not beenlooked at.

For this calculation we start at the motor, change that speed

with the transmission and final drive. The vehicle speed /

motor speed is locked by these ratios.

Motor RPM 1st Gear MPH 2nd Gear MPH 3nd Gear MPH 4th Gear MPH 5th Gear MPH

500.0 2.8 5.2 7.3 9.8 13.1

1000.0 5.6 10.3 14.6 19.5 26.2

2000.0 11.2 20.7 29.2 39.1 52.3

3000.0 16.9 31.0 43.8 58.6 78.5

4000.0 22.5 41.3 58.4 78.2 104.6

5000.0 28.1 51.6 73.0 97.7

6000.0 33.7 62.0 87.5

7000.0 39.4 72.3



Vehicle Speed

If we look at the 9 inch Advance DC motor running insecond gear we see a reasonable speed range for acommuter car with no shifting. This motor can reach6,000 RPM so this is a good fit.

The 9 inch Advance DC motor torque drops to about 50 Ft-Lb at 3,000 RPM. Torque continues to drop as speedincreases. That reducing torque at increasing speedreduces acceleration as speed increases. At 30 MPHthe 6.8 MPH/sec acceleration we started with has

dropped to 2 MPH/sec acceleration . At some point thespeed and torque will be in balance and there will be nomore acceleration.



Vehicle Speed

For the AC drive and 90 Ft-Lb to 4,000 RPM, I

need to change gears. In first gear I will get my

expected 4 MPH/sec acceleration, but I won¶t

have it in second gear. In second gear I will getabout 2.5 MPH/sec acceleration and in 3rd gear

it will be 1.8 MPH/sec acceleration. I doubt I will

ever need the higher gears.



Torque at the motor Constant Velocity

From the constant velocity note 123 Ft-Lb is

needed for 45 MPH at the wheels.

My final drive is 3.87:1 ratio so the input torque is

123 / 3.87 = 32 Ft-Lb.


32 1 3.58 9

32 2 1.95 16

32 3 1.38 23

32 4 1.03 31

32 5 0.77 41




A 9 inch Advance DC motor has 240 Ft-Lb peak torque at

low speed. As the RPM increase the torque available

decreases.

These input torques are well within the torque produced by

this motor at higher speeds for driving at 45 MPH

Output Torque Gear Ratio Input Torque Motor RPM Motor Torque

32 1 3.58 9 8040

32 2 1.95 16 4350 25

32 3 1.38 23 3085 4932 4 1.03 31 2300 110

32 5 0.77 41 1720 150

Motor torque must exceed the Input Torque




If the vehicle speed increase to 60 MPH more torque andmore motor speed is needed.

The required torque needed to maintain these speeds is notavailable from the motor in 2nd or 3rd gears.

Even the DC motor powered vehicle could benefit from attransmission at higher speeds.

Output Torque Gear Ratio Input Torque Motor RPM Motor Torque

40 1 3.58 11 10700

40 2 1.95 21 5760 11

40 3 1.38 29 4110 25

40 4 1.03 39 3072 49

40 5 0.77 52 2300 110




The AC motor has a continuous output of 60 Ft-Lbto 4,000 RPM and drops to 30 Ft-Lb at 8,000

RPM.Based on the previous torques the AC motor

could power the vehicle at 60 MPH in 2nd, 3rd or 4th gear.



Your Project

� The information presented here is basic physics and usesinformation from my conversion project collected from theinternet. It may not exactly apply to your project.

� You need to gather the data for you project and work

through it all.� Do calculations at different speeds.

� Do not ignore any of the data.

� These calculation should help avoid a conversion that haspoor acceleration, can¶t go up a hill, or can¶t keep up with

traffic.� There are still a multitude of things that can rob performancebut this will take care of the elementary mechanics.

� You are in charge and responsible for what you design andbuild. Please do it responsibly.

Basic Motion Calculations

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