My immediate thought is FAIL.
Same bike, same brakes, same everything, but bike A weighs 450 lbs., add another 100 lbs. to bike B @ 550 lbs.
Both bikes stopping from 60-0; Bike B will travel further. Not sure how to reason yourself out of that one.
motorcycle weight vs. stopping distance
- Thread starter kml
- Start date
My immediate thought is FAIL.
Same bike, same brakes, same everything, but bike A weighs 450 lbs., add another 100 lbs. to bike B @ 550 lbs.
Both bikes stopping from 60-0; Bike B will travel further. Not sure how to reason yourself out of that one.
HT_USMC
Registered
Grumble... now I've got to solve for V and A (at the very least) since the article doesn't spell out what those are!
Nice find Mr Biscuit. I'll attempt to do the math after all.
Just as a heads up, Alsterbator, I wasn't conceding my point I was trying to be gracious and rhetorical.
Also, having worked with tire manufacturer engineers I was informed maximum breaking for a tire is when they are locked up...i.e. locking up the brakes will stop you quicker than ABS. More control w/ ABS but further stopping distance.
As for motorcycles however we all know that locking up the front tire is not an option.
As for motorcycles however we all know that locking up the front tire is not an option.
I know this to be patently wrong.... stopping distances can close to double when a tire is locked up...
OK without getting all Newtonion and quoting said newtons laws I think this could be easily tested. Take a carboard box, put 10 pounds in it and accelerate it to a given speed on the living room carpet. Triple the weight and accellerate it to the same speed. Since we are talking about a maximum breaking event, we can just assume that the friction of the carpet to the box is the maximum breaking force that can be applied. Which goes farther? My money is on the heavier one.
Busa_Nick
Registered
Alsterbator
Registered
That says what I've been saying, Thanks!
Someone elses post.
It's plain english... It takes more brake pressure to stop a heavier weight, the added weight enables the use of more brake pressure due to the increase of traction
Alsterbator
Registered
Alsterbator
Registered
Just as a heads up, Alsterbator, I wasn't conceding my point I was trying to be gracious and rhetorical.
I never thought you did, I was agreeing to the variables comment.
Traction may increase a little but your contact patch doesnt. So although you may see an increase in traction, I doubt it is even remotely enough to offset the weight increase. Who wants to do the math of the deflection on this. Its not the same as getting direct downward force like the wing of a dragster. The momentum is carrying the weight foward, not down. Since your contact patch doesnt increase (significantly) the the possibility of overloading the friction of the contact patch increases with weight. Dont believe me, test for yourself. Take a quarter and a dime. Place them on a table. Lift one side of the table and see which slides first. My money is on the heavier quarter. Even though the quarter has a much greater contact patch, its weight will overload friction before the dime.
I always thought the same myself...I'm simply relaying what an engineer for a major tire manufacturer told me. If some guy gets paid the jack he did tells me different and holds the position he did I believe him, but I'm always open to being proven wrong. We're not talking about stopping on ice or a rainy surface...clean dry pavement he said. Lots of talk about friction and rolling friction.
Alsterbator
Registered
The suspension on a motor vehicle transfers the forward momentum into a downward force.
MelodicMetalGod
Registered
For the added weight to not be a factor in stopping distance, we must assume that tire traction will increase INFINITELY and in EQUAL or GREATER proportion to the added weight. Now I don't know for sure about what proportion tire traction increases (my guess is it's proportionally LESS than the added weight), but I'm pretty sure that every tire has a traction limit.
kml
Donating Member
That says what I've been saying, Thanks!
Someone elses post.
It's plain english... It takes more brake pressure to stop a heavier weight, the added weight enables the use of more brake pressure due to the increase of traction
So are we saying the additonal weight allows for a greater "maximum" breaking force, thus allowing to stop within the same distance? I thought everything, including the amount of breaking force applied, was equal.
I guess this would have a lot to do with the specific amount of addtional weight wouldn't it? There has to be a tipping point where the additional weight would outweigh (pun intended) the additonal maximum breaking force that would be allowed by said weight.
Alsterbator
Registered
Me on my bike weighing 200lbs@60mph can stop in the same distance as I can weighing 250lbsfrom 60mph.
Yes but that force applies additional load on the front tire but the additonal mass is also moving forward.
Similar threads
