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Given the equation S1 = S0 + V0*t + 1/2 a * t *t where S0 is inititial position (0 ft), S1 is current position (60 ft), V0 is initial velocity (0 ft/sec), t is time in seconds and a is acceleration, you can reduce the equation and solve for a = 120 / (t*t). Then since the acceleration of gravity is 32 ft per second squared, g = a / 32 or 3.75 / (t * t). This only works if you assume that the vehicle accelerates at a constant rate though...
 

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Discussion Starter · #5 ·
Thanks for the link and the math explanation..
According to the link Im pulling 1.41 ave G's with my 1.631 60ft.
It also stated that Im doing 50.2 MPH at the 60ft mark...
 

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Bob, Yea, you have to register and log in and use the link, but its free. I havent been on there for awhile and dont remember my login info:clonk:

But I played around with it a long time ago, If I remember right, 1.2 60ft is 2g's and 60 mph was around a 1.4 something. That means 0 to 60 in about 2 seconds for a lot of us. Kinda makes all those old car adds where they brag about their cars going from 0 to 60 in 6 seconds seem lame.

Scott
 

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You can also find this out the Mr. Wizard way.

Attached a protractor (used for measuring angles) in your car, with the flat edge on a horizontal "bubble" (i.e. level).
Hang a weight of known mass from the center zero-angle point. As you lunch, the maximum angle the string travels from zero will allow you to calculate maximum acceleration and likewise maximum G's. Or, just buy a G-tech. ;)
 

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That is one cool site:thumbsup: thanks for posting it.
 
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