Remember your HS Physics?

Love the responses, guys! My original post was prompted by the video of the DP that rolled. The issue of his speed at the time came up. I was suggesting that the faster one drives, the more severe an accident is likely to be.

johnnyrv wrote:
motorweek on pbs rates cars on their stopping distance which varies widely from car to car. so the real world is more complex the math models of simple physics.

What makes a major difference is that total stopping distance is from the time a computer activates the brakes to zero feet per hour.

NOT the same as Wheel Lock

ON some vehicles the ABS keeps the wheels just shy of lock.. This is supposed to be the best breaking point.... Math here is much, Much more complex.

My figures are SKID distance... From the point of wheel lock.

Old-Biscuit wrote:

Hope you don't mind, I saved that one and put it in my photobucket for future use!!! Chris

"When the wheels lock" I didn't see that and that takes it out of the realm of real world useable and useful data, because the last thing you want to do is lock the brakes. And your data explains why.

motorweek on pbs rates cars on their stopping distance which varies widely from car to car. so the real world is more complex the math models of simple physics.

fj12ryder wrote:
Pretty funny explanation. :) Helps to explain why motorcycles can outbrake a semi.

Actually they can not.. There are a couple other factors I did not leave out but did not mention.

I said "FROM THE TIME THE WHEELS LOCK"

That is where the smaller vehicle has the advantage,,,Wheels lock faster. Physics from there on is the same for all.

Also conditon of tires on dry pavement does not matter.

Much of this has been confirmed by interview with Police Department Accident Investigators and Re-constructionists (one of whom used to be a K-9 Officer, I often petted his partner).

They use the same skid formulas for ALL vehicles without Anti-lock
and the same set for all vehicles WITH Anti-lock.. Only anti-lock makes a difference.. and I'm not sure what that difference is.

But those are based on skid marks, which do not happen till the wheels lock up.

fj12ryder wrote:

wa8yxm wrote:

GordonThree wrote:
So the surface area and number of tires, brake pads, rotors, drums, etc play no part in your stopping distance?

That is correct.. I will explain why

The force of stopping is the weight of the vehicle (mass times gravity)

(Gravity is a constant for all vehicles) Times the coefficient of friction Thus the heavier the ride the more force stopping.

Braking power if you want to factor in area of tires in contact with road then the amount of force on each square inch of tire is

Weight/Number of square inches

And the stopping power is Friction (Pressure times coefficient) times square inches

So to completly make up a non-existent vehicle let's say we have a 10,000 pound ride with 100 square inches on the road

That's 100 pounds per square inch... Stopping power is 10,000C

now let's increase the number of tires so we have 200 square inches

We are now running 50 psi.. And the stopping power is 200$c or 10,000C (Where C is that coefficient I spoke of)

What happens if we cut the weight in half?

100 Square inches we now have 50 PSI and 5,000 C
200 Square inches it is 24 PSI and again 5,000 C

This is why vehicle size, number of tires and such do not matter.

Pretty funny explanation. :) Helps to explain why motorcycles can outbrake a semi.

I repeated tire tests different brands will stop the same vehicle in a much shorter distance that other tires. Real world always ends up messing with nice pure calculations

Sport45 wrote:
That's true if the acceleration is constant.

That's right. Otherwise one needs the velocity curve verses time and then do a little calculus.

wa8yxm wrote:

GordonThree wrote:
So the surface area and number of tires, brake pads, rotors, drums, etc play no part in your stopping distance?

That is correct.. I will explain why

The force of stopping is the weight of the vehicle (mass times gravity)

(Gravity is a constant for all vehicles) Times the coefficient of friction Thus the heavier the ride the more force stopping.

Braking power if you want to factor in area of tires in contact with road then the amount of force on each square inch of tire is

Weight/Number of square inches

And the stopping power is Friction (Pressure times coefficient) times square inches

So to completly make up a non-existent vehicle let's say we have a 10,000 pound ride with 100 square inches on the road

That's 100 pounds per square inch... Stopping power is 10,000C

now let's increase the number of tires so we have 200 square inches

We are now running 50 psi.. And the stopping power is 200$c or 10,000C (Where C is that coefficient I spoke of)

What happens if we cut the weight in half?

100 Square inches we now have 50 PSI and 5,000 C
200 Square inches it is 24 PSI and again 5,000 C

This is why vehicle size, number of tires and such do not matter.

Pretty funny explanation. :) Helps to explain why motorcycles can outbrake a semi.

I didn't read all 5 pages of this thread but, there is a line from a old song that goes called "Kodachrome" that goes something like ,

"When I think back on all the******I've learned in high school
It's a wonder I can think at all
And though my lack of education hasn't hurt me none
I can read the writing on the wall"