Stopping Distance

[TiGeo]

How much does my car weigh?

 

[MolaKule]

How much does my car weigh?

View attachment 320395

In the stopping distance equation of post #1, mass falls out so weight is irrelevant.

 

[TiGeo]

In the stopping distance equation of post #1, mass falls out so weight is irrelevant.

Ok got it now. So what about the friction coefficient...that's the unknown then in this case where we have speed and distance. I would think that should vary on tire compound (higher than 0.7 for these performance tires). Plugging in 0.7 in the formula souldn't solve the equation with those other variables?

 

[MolaKule]

Ok got it now. So what about the friction coefficient...that's the unknown then in this case where we have speed and distance. I would think that should vary on tire compound (higher than 0.7 for these performance tires). Plugging in 0.7 in the formula souldn't solve the equation with those other variables?

In this problem, u = 0.7 was given.

 

[TiGeo]

In this problem, u = 0.7 was given.

No. I'm asking if you take the stopping data I provided in my post and plugged it in to the equation what would the friction coefficient be? And if you used 0.7 with my stopping data the equation likely won't work.

 

[MolaKule]

No. I'm asking if you take the stopping data I provided in my post and plugged it in to the equation what would the friction coefficient be? And if you used 0.7 with my stopping data the equation likely won't work.

Your questions was, what is the weight of my car with no indication of what equations this was based upon.

Rearranging the distance equation given in post 1, take your stopping distance d and use
u = V^2/2gd to get a new value of u.

My value of u given was an average value for dry pavement and no skidding.

 

[TiGeo]

Ok, so I get for my best 60 to 0 panic stop w/ABS data:

(26.82)^2/2*9.8*32.45 = 1.1 as u which makes sense/jives with what I found for high-grip performance tires

 

[MolaKule]

Ok, so I get for my best 60 to 0 panic stop w/ABS data:

(26.82)^2/2*9.8*32.45 = 1.1 as u which makes sense/jives with what I found for high-grip performance tires

See post #14 for more detailed equations which include driver reaction times, etc.

ABS also has a finite reaction time as well.

 

[TiGeo]

See post #14 for more detailed equations which include driver reaction times, etc.

ABS also has a finite reaction time as well.

Got it.

For this, the Dragy (GPS/app for timing drag racing, track, etc.) starts when you slow down past 60mph so here I was on the brakes before getting to 60 so effectively a 0 reaction time when it started recording. I was just curious about the friction coefficient and my data supports what I found online about peformance tire values - 0.8 for a good street tire to over 1 for performance tires with slicks obviously higher. I've wanted to test my 200tw track tires in this manner, maybe I'll do it when it's warmer to compare. Can't test on track b/c you can't come to a complete stop but on the street should show an improvement over my current distance on Continental ECS02s even if the track tires aren't at full temp.

 

[Barkleymut]

If the road is wet and there is a child in the back seat, Goodyears will stop 20 feet shorter than other brands....i saw that in a commercial so it must be proved by math.

 

[rstsco]

The simplest equation for stopping distance of a vehicle is

Workdue to friction = - umgd = - 1/2 mV^2.

u - friction coefficient, m is mass of the vehicle (kg), v is the vehicle's speed (m/s), and g is gravitational acceleration = 9.8m/s^2.

So solving for stopping distance d = V^2/(2ug).

A car that was moving at 50 mph is stopping on a dry, flat surface in which the tire/surface friction coefficient is u = 0.7 without skidding.

What is the stopping distance in feet?

Thank you for intermittently posting technical questions on various topics. It's a nice distraction.