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i was just wondering if any of you good folks may know at least a ballpark figure of how many feet it would take in a panic stop from about 15mph in my 99 quad cab long bed unladen with rear drums and relatively new front pads & rotors. i have already tried google without any luck...thank you in advance for any assistance
 

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about 11 feet, if ya need the math I'll post it (assuming dry asphalt, concrete would be a little less)

Source: I'm a civil/highway engineer :thumbsup:
 

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Assumptions:
-dry road
-good tires
-reaction time not considered
-braking efficiency = 85% (would be 100% for anti-lock brakes)
-deceleration rate = 11.2ft/s2 (average rate for average vehicle on the road)
-roadway grade = 0%

S = initial velocity ^2 / 2 * gravitational constant (braking efficiency * coeff. of friction)
S = 22ft/s ^2 / 2 * 32.2 ft/s^2 (0.85 * 0.8)
S = 11.05 ft

I could do a calc that incorporates perception/reaction time, roadway grade, aerodynamic resistance, exact vehicle mass, etc. but it’s a pretty lengthy derivation and calculation with a ton of factors.

Just be in mind that this is only 15mph, if you’re doing 80mph, the distance required (based on this equation) is over 300 ft.
 

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.....
-deceleration rate = 11.2ft/s2 (average rate for average vehicle on the road)
Cool math. Guess it's a decent ballpark figure, but...............
Does the average deceleration rate for the average vehicle on the road really cover a 4 ton truck?

RJ
 
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