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4,409 Posts

The formula is

speed=square root of 30 X Distance X f

f=coefficient of drag

f=F/W

F=force in pounds to move drag sled (need drag sled and weight scale)

W=Weight in pounds of drag sled

Most crashes you have to use average skid distances. Measure skidmarks and divide by number of skids. This would be Distance. This is if you have 100% brake efficiency. If not then you have to adjust

speed=square root of 30XdistanceXfXn

n=braking percentage

So if you have a front wheel drive that had all four wheels lockup you have 100% so n=1

If you had the front right wheel not working you would have lost 35% of your total effectiveness, therefore n=.65

I hope i didn't confuse you too much!

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6,402 Posts

Ok the way I see it 121 feet dry road, approximately 50 or so MPH.

Probably would have been longer but a brick foundation has some stopping power...

Thanks for the formula, I did not want to drag out my notebooks from the academy

G

energy_dissipated_in_skid = brake_force * distance

Set these two equations equal to each other, then solve for speed:

speed = sqrt( 2 * force * distance / mass )

In a 1.0 g panic stop (

brake_force = mass * gravity

[Aside: if the braking can't reach 1 g, then the brake_force will be some factor less than mass * gravity; e.g. brake_force = 0.9*mass*gravity for a 0.9 g panic stop). Gravity is a constant based on how much the earth pulls on you: 32.2 ft/(sec^2) or 9.81 meters/(sec^2)

So the speed equation reduces to:

speed = sqrt( 2 * gravity * distance )

g = 9.81 meters/(sec^2)

and in your example, distance is 121 feet (36.9 meters), so

speed = sqrt ( 2 * 9.81 * 36.9 ) = 26.9 m/s or 60.1 mph

There are many little items not shown in the above: the energy stored in the rotating mass (wheels, tires), the number of wheels involved in the stop, and the change of front-to-rear wheel loading due to the decceleration acting on the center-of-mass of the car. The devil in those details aren't worth posting here.

H50's post has similar math, I'm just showing where it comes from.

HTH. - Ken

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