# Estimating vehicle horsepower and torque

• calan
In summary: The conversion factor is 0.74, so the result is 604 N m, which is about 445 ft lb. It looks like 604 ft lb is correct.I only see one problem. You are given the rolling resistance coefficient Cr, but you use Cr = 0. If you use Cr = 0.015, then the result is Fr = 4.7, which is more reasonable. In summary, the equation to estimate instantaneous engine torque (or horsepower) at the wheels is F = (m * A) + (.5 * Cd * FA * d * V * V) + (G * Cr * m). The net force on the vehicle can be determined by knowing the acceleration, velocity,

#### calan

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:

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.

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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.

voko said:
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

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Yes, that seems correct.

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.

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.

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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.

## 1. How is horsepower and torque measured in a vehicle?

Horsepower and torque are measured using a dynamometer, which is a tool that measures the power output of an engine. The engine is connected to the dynamometer, and as the engine runs, it creates a force on the dynamometer's rotating drum. This force is then used to calculate the horsepower and torque of the engine.

## 2. Can horsepower and torque be increased in a vehicle?

Yes, horsepower and torque can be increased in a vehicle through various modifications such as adding a turbocharger, upgrading the exhaust system, or tuning the engine. However, it is important to note that these modifications may also affect the overall performance and reliability of the vehicle.

## 3. What is the difference between horsepower and torque?

Horsepower and torque are two different measurements of a vehicle's power. Horsepower is a measure of how quickly an engine can do work, while torque is a measure of the twisting force produced by the engine. In simpler terms, horsepower is the "speed" of the engine, while torque is the "strength".

## 4. How do factors such as weight and aerodynamics affect horsepower and torque?

Weight and aerodynamics can have a significant impact on the horsepower and torque of a vehicle. A heavier vehicle will require more power to move, reducing both horsepower and torque. Similarly, poor aerodynamics can create drag, which also decreases the power output of the engine.

## 5. Are there any limitations to estimating horsepower and torque in a vehicle?

While dynamometers can provide accurate measurements of horsepower and torque, there are limitations to estimating these values in a vehicle. The results may vary depending on the testing conditions, such as temperature and humidity, and the accuracy of the dynamometer itself. Additionally, external factors such as tire pressure and road conditions can also affect the measurements.