Horsepower vs hill gradient problem

[donperry]

So, I have a car I'm tuning using a road dyno software. This software basically uses the gear ratio of the car, it's total weight, and the speed it moves the RPM to determine it's power output (HP and Torque)

In order to use this software a flat road is needed. That works well and is very accurate compared to real dynos

My area doesn't have a road that long for tuning in higher gears, But there is a very long hill with a fixed gradient of 12.8%. If the test is done on this hill, unsurprisingly, the output from the software is much lower. How can I use this information to correct the output from the software.

EGill of 12.8% grade reads 280hp
What is the true power if the road was flat?

 

[jerromyjon]

Something doesn't seem right, unless the car is very high horsepower to weight ratio.

 

[russ_watters]

Welcome to PF!

You could use E=mgh to find the additional energy added during the climb. Then you have to divide it by time to climb to get the average added power. The problem is, if the speed isn't constant, you'll have to adjust for that, which isn't necessarily straightforward...

Also, do you know how this software does its calc? Is it timing the car's acceleration?

 

[CWatters]

Also, do you know how this software does its calc? Is it timing the car's acceleration?

I think that's what he means by

the speed it moves the RPM

That sounds like it's calculating angular acceleration as in...

Power = Torque * angular acceleration

Edit: opps that's wrong.

 

[donperry]

Thanks for the replies.

Yes, this software uses some formula against the speed of which the RPM climbs,
Each sample records the current RPM taken at that time


For example,
With gear ratios and wheel size factored in it then figures how much energy it would take to move the RPM of a vehicle up by 50 rpm (for example) in just 20 milliseconds with a total mass of 1500KG ,

 

[CWatters]

The Kinetic Energy of the car is

KE = 0.5mV²
where
m=mass
V=velocity.

The KE gain is..

ΔKE= 0.5m (V_f² - V_i²)
where
V_f is the final velocity
V_i is the initial velovity

So if you only know the rpm increase you can't calculate the energy required. That will vary depending on the initial velocity. eg it takes more energy to accelerate from 20 to 30mph than it does from 10 to 20 mph. Even though the velocity increase is the same in both cases.

You have to convert the initial and final rpm to initial and final velocity (using the gear ratio) before you do the subtraction to calculate the "increase" in rpm or velocity.

 

[donperry]

You can. The RPM is used to calculate the vehicle speed. They already know the mass.
They also factored in coefficient of drag and frontal area.

I was wrong for using "rpm change"

 

[CWatters]

Energy is only used to overcome drag (in all its forms) and to accelerate the mass. Energy is not required to move the mass at a constant speed (if you ignore drag). So my guess is they only use the vehicle speed to calculate the drag.

The extra power required to climb a hill at constant velocity V_C is given by..

Power = force * velocity
where in this case..
force = mass * acceleration due to gravity (aka "g")
velocity = V_V = the vertical component of the cars velocity

V_V = V_C*Sin(theta)
where
"theta" is the angle of the hill and
V_C is the velocity of the car

So putting that all together the extra power is..

Power = mass * g * V_C * Sin(theta)

Example:
The extra power required by a 1000Kg car going up a 10 degree slope at 5 meters per second...

= 1000 * 9.81 * 5 * Sin(10)
= 8517W
or
8517/750 = 11HP

 

[donperry]

thanks for that