motorcycle weight vs. stopping distance
- Thread starter kml
- Start date
KokomoBusa
Banned
like i said, study harder jr....hate to see you throw away 7 years like that. to think that objects falling due to gravity and the inertia of two objects with different weight is even remotely similar demonstrates something called GROSS CONCEPTUAL ERROR.
To me its simple. The op said max breaking event. If your contact patch is at max load, you will not be able to apply any more breaking force. If you cant apply more breaking, how are you going to DOUBLE your breaking effort. You have to double the breaking in relation to double the weight. Add in G forces doubling your weight the required breaking force likely grows exponentially. My opinion is that someof you guys are stuck on a general principal that works within a LIMITED envelope trying to apply it to a specific type of breaking, max breaking. I can stop my bike in 500yards alone or 2 up. I Cant do the same at max breaking in the shortest possible distance. Like I said if Alster were correct the inverse would have to be true as well. There would be no diff between a heavy rider and a skinny rider for acceleration but we know thats not true.
Robot
Donating Member
OH, LORDY..........Sorry Physics is my thing, and passion. There is some good and some bad physics in this thread. I have pushed the pencil through MANY similar problems. I will do my best to set some things straight.Physics Forum, Wikipedia, etc. often will give you the wrong/incomplete information. Earlier in the thread Mu sub S, and Mu sub K were given for static and kinetic(moving) friction calculations. If you analyze these, (not directly applicable),calculations you will find Mu sub S is NOT dependant upon surface area contact. WOW!! , this means a glass setting on the small surface area rim, has the same friction as when it sets on it's (large surface area base).
This means a dragster would have the same traction with bicycle tires, why do we bother putting these crazy wide slicks on the back!
Because Drag racers know better, ALL general Mu sub S and Mu sub K friction calculations ASSUME surfaces are (Perfectly flat) 2 dimensional surfaces. This works pretty well with the smooth glass on the formica countertop. BUT, real world tires are VERY different and VERY 3 dimensional in surface contact,(shearing forces must now be taken into account). That's why these equations,(without modification), dont work for this application.
Back to the original question.........
It is a TRICK QUESTION, what we are talking about is a moving object that we alter with a passenger, (increasing mass). The Bottom line initially is about conversion of kinetic energy to thermal energy. We increase kinetic energy by increasing the bike's effective mass. Formula for Kinetic Energy is 1/2 Mass times the Velocity squared.
The good news is that when we look at (K.E. = 1/2M x V^2) is that the 1/2 in front of the Mass is, (from a math perspective), not very powerful. We can increase the mass some and not have a profound effect. The other side of the coin is the squared term on the Velocity, a small increase in velocity has serious kinetic energy influence.
Blah Blah Blah.....now for the conclusion
DONT SPEED WITH A HEAVY PASSENGER ON YOUR BIKE
The passenger has INCREASED kinetic energy, (some passengers increase more than others
). This increase in kinetic energy must be thermally dissapated through the brakes, tires, etc. Does this mean more stopping distance? maybe/maybe not. Extra effort WILL be needed to keep the same stopping distance BUT, if you dont exceed capability of brakes/tires you CAN stop in the same distance. (Yes Mu sub S and shearing forces will increase some from the added weight, this seemingly simple problem is actually very complex).The problem varies even more with added Velocity, that increase in mass,(passenger), makes things FAR worse at higher speeds.
Hopefully this clears some things up, pm me if you have questions. Sorry if I rant when it comes to Physics(or bad Physics). Yep, I'm a nerd.
eace:If your still bored try to answer this one:
I have seen shock sensors (little sand filled simple mechanical devices) on the outside of some shock sensitive containers (packages). These can be purchased in various "G" ratings. Lets say I have a shock sensitive package(5 lbs.) and I drop it from my waist to the concrete floor, (3 feet), how many g's will show on my indicator?
Alsterbator
Registered
Thats one of the very reasons to understand GVWR, the braking systems on any motorvehicle has to be able manage the braking forces at full load... or Max Weight. Keep in mind that what locks the front wheel up is not the progression of braking forces, it's the quickness in which it is applied.
Thats the very reason why they stress "progressive squeeze of the front brakes, the slower you get the more pressure that can be applied", If you "grab" (squeeze too quickly) the front brake then you exceed the tires ability to maintain traction (rider error not tire failure).
Motorcycle "A" with added weight (still under GVWR) will be able to apply more brake pressure progressively in a quicker manner than Motorcycle "A" (same motorcycle) with less weight.
Robot
Donating Member
Sounds good to me
My real point,(from a Physics calculation perspective), is that "some" increase in mass at "low" speeds will have little effect, But Speed is critical here, the higher the speed, the more significant the impact ,(on braking), from that "mass increase".
kml
Donating Member
Sounds good to me
My real point,(from a Physics calculation perspective), is that "some" increase in mass at "low" speeds will have little effect, But Speed is critical here, the higher the speed, the more significant the impact ,(on braking), from that "mass increase".
Will a 10 lbs bag of potatos on the passenger seat, have no effect at 30 mph but add a couple of feet at 160 mph?
cheers
ken
Robot
Donating Member
Yep the bag WILL have an effect at 30 MPH, most likely soooo small that most riders could'nt detect it though. The Exact effect would need solid numbers for the calculations.
Energy difference between 30 and 160 is profound. Again solid numbers needed for calculations, but, my gut says more than a couple feet at 160.
Think of the energy if you got hit by a 10 lb. bag of anything at 160. I'm over 200lb. but I bet it would knock me on my butt.
Commuta_Busa
Donating Member
I guess I'm to late with the right answer. Looks like this has been talked out already.
My answer was... it depends.
My answer was... it depends.
Robot
Donating Member
I guess I'm to late with the right answer. Looks like this has been talked out already.
My answer was... it depends.
Never too late for the "most direct" correct answer
Commuta_Busa
Donating Member
MelodicMetalGod
Registered
Robot,
Glad to see a true physics "nerd" join the fray. I dig physics concepts, but I don't care for the tedium of formulas and calculations. Thanks for your contribution!
Glad to see a true physics "nerd" join the fray. I dig physics concepts, but I don't care for the tedium of formulas and calculations. Thanks for your contribution!
jellyrug
Donating Member
In practice this is almost impossible to calculate with accuracy, as the package will accept some of the impact and kinetic energy will transmit into package deformation. In theory package deformation will be the distance travelled after impact.
We do not need the weight, for the same reason that in pure applied math theory weight makes no difference to stopping distance, the initial topic of this thread.
In theory we need to know either the very small distance the package deforms after impact, or the time to calculate the g's.
Let's assume it is 1/4"
3 feet = 0.9144m
1/4" = 6.35mm
Velocity before impact:
v = SQRT(2gh) = SQRT(2x9.806x0.9144) = 4.2348m/s
Acceleration:
a = (v^2-u^2)/2s
a = (0^2-4.2348^2)/2/0.0064 = -1412m/s^2
g's = acceleration/gravitational acceleration
g's = -1412/9.806 = -144 g's
We can also calculate the force of impact with the weight you provided:
5lbs = 2.2727kg
Force of impact = mass x deccelaration
F = 2.2727 x 1412 = 3209N
For the imperialists that would be 3209/9.806 x 2.2 = 720 lbs force
I'm not a Physicist like you, use to be an Engineer a long time ago, before I needed to earn more money. Therefore I am a little rusty, can't remember this was either final school year, or first year college. So, how did I do?
jellyrug
Donating Member
All things being equal, the light bike will stop quicker.
I've never seen such a collection of would be professors and bullsh*t on vehicle dynamics in my life...
Blue
Is that gut feel, guessing, or can you substantiate with a few facts? That is about stopping quicker, not the rest.
There is no simple answer to this question because everyone wants to prove everyone else wrong. That means we want to throw monkey wrench type examples into the works.
I would bet a rear wave rotor that if you asked the leading physicists of the world the following question you would get the answer below:
Question: Will two of the same motorcycles with different weight riders stop in the same distance with the following assumptions:
1. Their weights are not so extreme they cannot physically ride the motorcycle.
2. The motorcycles have normal stock tires at generally stock pressures.
3. The motorcycles are generally the same.
4. Either rider can, but will not, lock the wheels under the braking test speeds.
5. Each rider is equally capable of braking a motorcycle.
Answer: Yes.
I would bet a rear wave rotor that if you asked the leading physicists of the world the following question you would get the answer below:
Question: Will two of the same motorcycles with different weight riders stop in the same distance with the following assumptions:
1. Their weights are not so extreme they cannot physically ride the motorcycle.
2. The motorcycles have normal stock tires at generally stock pressures.
3. The motorcycles are generally the same.
4. Either rider can, but will not, lock the wheels under the braking test speeds.
5. Each rider is equally capable of braking a motorcycle.
Answer: Yes.
05 Busa LE
Donating Member
kml
Donating Member
Alsterbator
Registered
Is that gut feel, guessing, or can you substantiate with a few facts? That is about stopping quicker, not the rest.
That is a gut feeling, he believes as most here do that the what he feels should be obvious or common sense is the right answer... He cannot back up his claim because it has never been truly tested. I guess what I have learned so far in this thread is that when riders are presented with a theory (that has been proven) and they cant comprehend how its plausible then it must be wrong.
This thread provided a great learning opportunity for everyone but most wanted to focus on discrediting the theory rather than try and understand it.
All things being equal... the only variable being weight, a motor vehicle is capable of stopping in the same distance loaded as it is unloaded... The rider is the limiting factor, based on the response in this thread I would say that it is infact the riders mentality (they don't believe they can, therefor they can't) that limits them.
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