# Vehicle Simulation Question

1. Feb 17, 2010

### Peetch

I have been reading on these forums for a bit learning about vehicles and their physics. I still have a few questions and do wanna clarify some things as well.

I am trying to simulate a vehicle give some basic info you can find online about any car or vehicle. I know Peak HP and the corresponding RPM, I know peak torque and the corresponding RPM and i know the idle RPM and HP. given that info i make a linear graph for the HP of the engine by using the HP = (Torque * RPM)/5252 and its variations. Once i have a close approximation of the HP curve given the vehicles RPM i can then calculate the engines output torque.

I know the gear ratios of the vehicle as well. and they are as follows:

Gear 1 = 3.49
Gear 2 = 1.86
Gear 3 = 1.41
Gear 4 = 1.00
Gear 5 = 0.75
Gear 6 = 0.65
Differential Gear = 1.7

now using EngineTorque * CurrentGear * Differential I will get the torque to the drive wheels.

The part I am a little lost on is if the vehicle is in second gear with the engine RPM at 3000 and using our liner graph the approximated engine HP is 141.47. With that we will take Torque = (141.47 * 5252)/3000 so our engine output torque will be 247.67. Now applying that through the transmission and so on the torque to the drive wheels will be 247.67 * 1.86 * 1.7 and that equals 783.13.

Now this is where I am kinda lost. I know that the torque going to the wheel is 783.13 but how does that apply with MOI and AA and the vehicles speed. Here is other data i know but not sure how to apply it.

Vehicle Weight: 5200 pounds

again this is my understanding if I an wrong pleas correct any of my steps and please help me understand how I go from torque to the wheel to the speed/movement of the vehicle

2. Feb 17, 2010

### Bob S

141.47 HP x 746 = 105,537 watts

torque = 105,537 watts/[2 pi x RPM/60] = 105,537 watts/[2 pi x 3000/60] =336 Newton meters (torque)

(Are you using non-metric units for torque?)

Torque at drive wheel in 2nd gear is 336 x 1.86 x 1.7 = 1062 Newton meters.

So maximum power-limited horizontal wheel force on pavement = 2167 Newtons.

In lower gears you are not power-limited but friction (torque)-limited at drive wheels. What is vehicle mass M, and what is percentage (pct) on drive wheels? The maximum horizontal accelerating force is

Fmax = pct x M x g x ~80% (for static friction limit).

[added] If your vehicle'e mass is 1000 Kg, its weight is 9810 Newtons.

The maximum force on roadway is 9810 x 50% x 80% = ~3924 Newtons.

Bob S

Last edited: Feb 17, 2010
3. Feb 17, 2010

### Peetch

the vehicle mass is 2359 kg so 23118.2 Newtons
the vehicle is a normal 4 wheel vehicle with about 50% of the mass is on the drive wheels.

Also how did you determine this?

4. Feb 17, 2010

### Bob S

2359 Kg is a very heavy vehicle. The weight on the drive wheels is ~ 11,570 newtons. The maximum static friction force on roadway w/o slipping is ~80 % of this, or 9260 Newtons. So it looks like you are power limited.

The kinetic energy of a moving vehicle is W = ½·M·v2

For constant power, the acceleration and velocity are

P = dW/dt = constant = M·v·dv/dt = M·v·a.

Bob S

5. Feb 17, 2010

### Peetch

Maybe you can better explain something to me, How do we convert from drive wheel torque to forward velocity of the vehicle. I am missing something and it isn't making a whole lot of sense.

6. Feb 17, 2010

### Bob S

In Post #2, the drive wheel torque was calculated to be 1062 Newton meters.

The drive wheel radius is 0.49 meters.

So the accelerating force at the perimeter of the wheel is 1062/0.49 = 2167 Newtons.

Remember that in general power (watts) P = 2 pi RPM/60 x torque

Wheel perimeter velocity = street velocity v = 2 pi x wheel RPM/60 x wheel radius (in meters per second)

Wheel RPM = engine RPM/3.16