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## Topic review (newest first)

alex6192
2005-09-06 11:07:50

Thanks very much, I've got it now

Thanks again

Alex

MathsIsFun
2005-09-05 10:51:30

Quick Answer: 0.24 × (22/15)² = 0.516

If you are happy thinking about a cars speed in feet per second, then just use the 0.24, but if you want to use mph, you need to use 0.516

Example using 66 feet per second (which is 45 mph):

F = 0.24 × (45 × (22/15))² = 0.24 × (66)² = 1045

But we expand this a little: F = 0.24 × (45 × (22/15))² = 0.24 × (45)² ×  (22/15))² = 0.24 × (22/15) × (22/15) × 45²
And let's just combine the fixed numbers: F = 0.24 × (22/15) × (22/15) × 45² = 0.516 × 45²

Let's check that we get the same answer:

F =  0.24 × (66)² = 1045
F =  0.516 × (45)² = 1045

Also check out Unit Conversion Tool

And this one is about Area Conversion, it just shows the concept of applying the conversion twice.

Alex6192
2005-09-05 07:58:48

Hi,

I am writing a phsyics engine for a car game. I am having problems understand a drag equation which I have found.

I am using the following resource.

http://www.gamedev.net/reference/articles/article1615.asp

The equation is:

Where:

C = coefficient of friction, a factor depending on the shape of a car and determined by experiment; for a late model Corvette it is about 0.30;

A = frontal area of the car; for a Corvette, it is about 20 square feet;

P = density of air = 0.0801

V = speed of the car.

If you plug in the values, you get:

Up to this point I understand.

Then the resource says the following:

"We want, at the end, to have V in miles per hour, but we need in feet per seconds for the calculations to come out right. We recall that there are 22 feet per second for every 15 miles per hour, giving us:"

What I dont understand is how they go from 0.24 to 0.517.

Any help would be great

Many thanks
Alex