Calculating road-load horsepower

[Guest Ash]

 

The following is part of a reply to a General Discussion thread. I thought I'd post the tech-stuff here for anyone who likes a play, and wants to know how to calculate the power their tyres are actually delivering into the road.

 

All you need is an accelerometer and the ability to perform basic maths. You will also have to find a flat road of a suitable length. Changes in slope, during the measurement, will induce error. Also, for greater accuracy, it is best to do several runs and average the readings.

 

The process determines what I call "spot horsepower". Which is horsepower at a particular chosen rpm. To plot a graph of power output (or torque) over a spread of rpm, it will be necessary to do repeated runs at, say, 500 rpm intervals. Then plot the results on a graph.

 

For the sake of this example, I will assume a spot-power reading at 5000 rpm.

 

Okay, so here's the procedure:

 

1) Pop along to a public weighbridge and weigh the car (noting the fuel load). Add the weight of the two occupants who will perform the tests (one does the driving and the other takes the readings).

 

You only need to do the above procedure once. Provided you keep the same fuel-load each time. Different occupants should weigh themselves and add to the weight of the car - which should be constant. Unless something has changed since it was last weighed, say, a different exhaust has been fitted.  

 

2) Find a long flat section of road. It also needs to be straight. You can often find sections of motorways that are ideal.

 

3) Choose a gear and begin accelerating from fairly low down in the rev range.  

 

4) Accelerate at maximum possible rate up to the chosen spot rpm. Which, for the sake of this example, is 5000 rpm.

 

5) At 5000 rpm the reading of the accelerometer and speedo is noted. Let's say, for the sake of this example, readings are 0.4 G at 100 mph.

 

NOTE: For greater accuracy I use portable data-logging equipment. But the stock speedo and rev counter can be used to good effect.

 

6) Continue accelerating for another few mph, then shift to neutral and coast along.

 

7) When the speed reduces to the same speed noted in (5) take a reading from the accelerometer. Here let's say the reading is 0.05 G

 

Note: This second reading tells you the load due to wind resistance, etc. Which is added to the accelerometer reading noted in (5).

 

So here is where the fun starts, the basic formulas are:

 

i) Force equals Mass multiplied by Acceleration, or F=MA

 

ii) Horsepower equals Force multiplied by Velocity, divided by 550.

 

Note: The 550 comes into play because one horsepower is 550 foot pounds per second. Expressed HP=(FV)/550.    

 

The example figures are:

 

Total weight = 3500 lbs; RPM = 5000; Speed = 100 mph; Maximum acceleration = 0.4 G; Coastdown deceleration = 0.05 G; Total acceleration (maximum + coastdown) = 0.45 G.  

 

First calculate the Force:

 

F = MA

 

Inputting our example figures gives, F = (3500 x 0.45 x 32.2) /32.2

 

Therefore, F = 1575 lbs

 

Note: You will see there are two extra figures in the above, of "32.2". If you do this yourself, simply insert your particular weight and acceleration figures leaving out the two "32.2's". They are performing a couple of conversions, which there's no real need to go into separately.  

 

Now, input the Force figure into the second formula:

 

Horsepower = (F V)/550.

 

Calculate Velocity in feet per second where V ft/sec = Vmph x 1.467. Our example speed is 100 mph or 146.7 feet per second.

 

Inputting example figures gives, HP = (1575 x 146.7) / 550

 

Therefore, HP = 420.09.

 

The engine torque output is derived using the formula: Torque equals 5250 times Horsepower, divided by RPM. T=(5250xHP)/RPM

 

Inputting our example figures gives T = (5250 x 420.09) / 5000

 

Therefore, T = 441.09 foot pounds.

 

It all sounds a bit fiddly at first. But with a little practice the process becomes automatic, and it's good fun too.

 

Yours,

J    

 

 

(Edited by Ash at 4:45 pm on Dec. 6, 2001)

[Adam W]

Quote: from Ash on 1:55 pm on Dec. 6, 2001[br]

 

 

First calculate the Force:

 

F = MA

 

Inputting our example figures gives, F = (3500 x 0.45 x 32.2) /32.2

 

Therefore, F = 1575 lbs

 

Note: You will see there are two extra figures in the above, of "32.2". If you do this yourself, simply insert your particular weight and acceleration figures leaving the two "32.2's" in place. They are performing a couple of conversions, which there's no real need to go into separately.  

 

 

Hang on a second!  F = (3500 x 0.45 x 32.2) /32.2

 

which can also be written as

 

F=  3500 x 0.45 x 32.2

    _____________

 

           32.2

 

 

Which is equal to 3500 x 0.45 isn't it?  Are you sure you got the brackets in the right place or have I missed something obvious?

[Guest Supragirl]

Quote: from Adam Wootten on 2:39 pm on Dec. 6, 2001[br

Hang on a second!  F = (3500 x 0.45 x 32.2) /32.2

 

which can also be written as

 

F=  3500 x 0.45 x 32.2

    _____________

 

           32.2

 

 

Which is equal to 3500 x 0.45 isn't it?  Are you sure you got the brackets in the right place or have I missed something obvious?

 

Well Adam if you have missed something obvious I think the rest of us have as well!!!!! lol lol

 

Or perhaps we don't have the "basic" understanding of formulae

 

 

(Edited by Supragirl at 3:45 pm on Dec. 6, 2001)

[Guest Ash]

 

 

Yes, Adam, you are quite correct. One cancels out the other. I inserted the two 32.2s for the benefit of the purists amongst us, that's all. They are performing a couple of conversions. But, as I did say, there is no real need to go into it.

 

Yours,

J

[Guest Ash]

 

 

Reference my correction under the General Discussion forum, I just re-read the post and I realised there was a grammatical error in the offending paragraph (which I have edited so it now reads correct).

 

I initially wrote the wording to describe the simple form of the expression F=MA. But then thought against it because, technically, the two 32.2s should be included for illustrative purposes. The wording was changed but I screwed up slightly.  

 

Sorry for the confusion.

 

Yours,

J

[Syed Shah]

Top post!

 

 

[Adam W]

Well, we'll forgive you this time Ash, but don't let it happen again, alright?

[Guest Ash]

 

 

LOL

 

Yours,

J