Estimating vehicle horsepower and torque

  • Thread starter calan
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  • #1
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Main Question or Discussion Point

Given the following instantaneous parameters and constants...

Known:

d = air density = 1.24
Cr = rolling resistance coeffecient = 0.015
FA = frontal area = 1.48
Cd = drag coefficient = 0.29F
G = gravity (mps2) = 9.80665
pi = 3.1415926

m = vehicle mass (Kg) = 1542
DL = drivetrain loss = .2

V = velocity (m/s)
A = acceleration (m/s/s)

Unknown:

F = vehicle force (KgMps2)
T = Torque (KgM)
HP = horsepower

Unknown, but can be found if needed:


Tire radius
Current gear and gear ratio


... what would be an equation that would let me estimate instantaneous engine torque (or HP) at the wheels?

So far (I think) I have worked out these terms:

rollDrag = G * Cr * m
airDrag = .5 * Cd * FA * d * V * V
F = (m * A) + airDrag + rollDrag

Assuming that ^ is correct, this is where I'm stuck.

Any help is greatly appreciated. I'm sure I'm just over-thinking it or something, but I've drawn a blank and have been spinning my wheels (so to speak) on this for a couple of days now.

Thanks in advance!
 
Last edited:

Answers and Replies

  • #2
6,054
390
Since you know the acceleration, you can determine the net force on the vehicle. Since you know the velocity, you can determine the drag force. Since you know the mass, you can determine the rolling resistance. Eliminating those, you get the propulsive force. The product of that and velocity is the power at the wheels. Knowing the drivetrain loss, you can obtain the engine power.
 
  • #3
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Since you know the acceleration, you can determine the net force on the vehicle. Since you know the velocity, you can determine the drag force. Since you know the mass, you can determine the rolling resistance. Eliminating those, you get the propulsive force. The product of that and velocity is the power at the wheels. Knowing the drivetrain loss, you can obtain the engine power.
So would this be correct?

F = (m * A)
airDrag = .5 * Cd * FA * d * V * V
rollDrag = G * Cr * m
Propulsive force = F + airDrag + rollDrag

power at wheel = Propulsive force * V (what would the units be here; metric HP?)
power at engine = power at wheel * DL
 
Last edited:
  • #4
6,054
390
Yes, that seems correct.
 
  • #5
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Hmmm... I've got some units messed up some where or something. I'm getting numbers that don't make any sense in my excel spreadsheet.
 
  • #6
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Ok... here's the numbers and math:

Code:
d = air density = 1.24
Cr = rolling resistance = 0.015
FA = frontal area = 1.48
Cd = drag coefficient = 0.29

pg = percent grade = 0
G = (m * pg) / SQRT(10000 + (pg * pg)) = 0

V = velocity = 37.8 mph = 16.9 m/s
A = acceleration = 4.1 mph/s * .4470 = 1.8168 m/s/s
m = vehicle mass = 1542 kg
DL = drive train loss = .2
GR = gear ratio = 1.115
TD = tire diameter = .632m 

 
F = (m * A)
F = 1542 kg * 1.8168 m/s/s = 2802 

Fd = .5 * Cd * FA * d * V * V
Fd = .5 * .29 * 1.48 * 1.24 * 16.9 m/s/s * 16.9 m/s/s = 76

Fr = G * Cr * m
Fr = 0

Fp = F + Fd + Fr
Fp = 2802 + 76 + 0 = 2878

T = (Fp * (TD / 2)) / GR 
T = (2878 * (.632 / 2)) / 1.115 = 815.65 N m
T ft lb = T *  .74 =  604 ft lbs
What am I missing? I'm pretty sure the car isn't putting out 604 ft lbs of torque to accelerate at 4.1 mph/s from 38mph. :smile:
 
Last edited:
  • #7
6,054
390
G used to be acceleration due to gravity, now it is something else and is zero, which gives you zero rolling resistance force. That can't be right.

Apart from this, the power is P = Fp * V = 2878 * 16.9 = 48638.2 W, which is about 65 hp, which does not look unreasonable. The angular speed of the wheels is w = V/R = 2V/TD = 2 * 16.9 / .632 = 53.5 1/s, and the engine speed is wE = GR * w = 1.115 * 53.5 = 59.7 1/s, and the torque T = P/wE = 48638.2 / 59.7 = 814.7 N m, which is close to what you obtain.
 

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