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Old 05-13-2013, 10:27 PM   #121
Tripped1
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Quote:
Originally Posted by bumbeen View Post
This statement confuses me. It makes me think that if I switch from a 400lb bike to an 800lb bike, I now need twice as sticky tires. If I am leaned over on a 400lb bike and I have the back tire in the air and all 400lbs on the front tire, how is that different from being on an 800lb bike and having 400lbs on the front tire?
It isn't functionally, that is why I said you don't want to be on the front tire while leaned. That being said a an 800 pound bike makes more centripetal force for any given speed vice a 400 pound bike. Since you pretty much have to estimate for 1G of acceleration, that means at that speed you either have to be going half of the speed or twice the radius to compare a 400# bike to an 800# bike.

That equation is F=(mv^2)/R in that equation mass is basically a coefficient.

So far as it goes, the downforce of a bike in a corner never changes, its equal to the normal, the bikes weight (assuming no hills) the only thing that changes are the demands on the contact patches, and remember these are vector sums, if at any given time the resultant equals more than 1G in any direction that isn't down the result is a slide. Since dynamic friction is less than static friction that is generally a bad thing.
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Old 05-13-2013, 10:40 PM   #122
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Originally Posted by Tripped1 View Post
No it won't, the traction to make that happen doesn't exist.
Five times the weight pressing down on the contact patches gives five times the friction (traction) which can exert five times the cornering force to a vehicle of five times the mass, so yes it does. Of course the comparison is silly because ten-ton sportscars with 2500hp don't exist and nor do the tires for them to ride on, but the principle is correct.

The reason formula one cars can pull so many g's in cornering is downforce. Due to their wings and other aerodynamic tricks, they have, say, three times the downward force on the wheels and therefore three times the traction (compared to a car with no downforce aerodynamics) but still the same weight and can therfore pull three times as many g's in a corner. Take a 500kg formula one car, remove the wings and add 1000kg of weight and it would have the same traction as one with wings but due to the greater mass, it would corner (about) the same as a 500kg formula one car with no wings.

Motorcycles experience no downforce so the difference in cornering ability between a 200kg bike and a 400kg bike would be zero if they were of equal size and steering geometry, etc. In practice of course, there are no 400kg sportbikes but there are 400kg cruisers and 200kg cruisers and I venture to say their cornering capabilities are pretty similar.

All of the more esoteric discussions about weight and traction on each wheel versus centre of mass and centripetal force are not too difficult for someone with a scientific/engineering mind to visualize but are damnably tricky to describe. I suspect that's why this discussion has taken off into science mumbo-jumbo land. There are also some people who can't get over their perceived (mis)beliefs around the difference between theory and practice, and finally there are those people who only THINK they understand the science. When you don't know what you don't know, and everyone's an internet expert, it's tough to tease out the real facts from the techno babble.
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Old 05-13-2013, 10:49 PM   #123
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Originally Posted by slartidbartfast View Post
Five times the weight pressing down on the contact patches gives five times the friction (traction) which can exert five times the cornering force to a vehicle of five times the mass, so yes it does. Of course the comparison is silly because ten-ton sportscars with 2500hp don't exist and nor do the tires for them to ride on, but the principle is correct.

The reason formula one cars can pull so many g's in cornering is downforce. Due to their wings and oter aerodynamic tricks, they have, say, three times the downward force on the wheels and therefore three times the traction (compared to a car with no downforce aerodynamics) but still the same weight and can therfore pull three times as many g's in a corner. Take a 500kg formula one car, remove the wings and add 1000kg of weight and it would have the same traction as one with wings but due to the greater mass, it would corner (about) the same as a 500kg formula one car with no wings.

Motorcycles experience no downforce so the difference in cornering ability between a 200kg bike and a 400kg bike would be zero if they were of equal size and steering geometry, etc. In practice of course, there are no 400kg sportbikes but there are 400kg cruisers and 200kg cruisers and I venture to say their cornering capabilities are pretty similar.

All of the more esoteric discussions about weight and traction on each wheel versus centre of mass and centripetal force are not too difficult for someone with a scientific/engineering mind to visualize but are damnably tricky to describe. I suspect that's why this discussion has taken off into science mumbo-jumbo land. There are also some people who can't get over their perceived (mis)beliefs around the difference between theory and practice, and finally there are those people who only THINK they understand the science. When you don't know what you don't know, and everyone's an internet expert, it's tough to tease out the real facts from the techno babble.

I didn't get into traction, because its beyond my realm. But no, half of the downforce with double the weight doesn't equal the same cornering ability. I was pushing vector sums for a reason, and while your example would certainly equal the same friction, it does not equal the same traction.

Note they are different terms.

A practical application of this is pretty much every racing vehicle on the planet, light is king, and its king for a reason.

.....and I'm not getting into dynamics modeling without getting paid or graded. Neither of which is happening here
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Old 05-14-2013, 12:17 AM   #124
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Quote:
Originally Posted by bumbeen View Post
This statement confuses me. It makes me think that if I switch from a 400lb bike to an 800lb bike, I now need twice as sticky tires. If I am leaned over on a 400lb bike and I have the back tire in the air and all 400lbs on the front tire, how is that different from being on an 800lb bike and having 400lbs on the front tire?
Sorry, I stated it poorly. I was speaking about weight transfer. In a simplistic way of imaging it think of your forks pushing on your front tire as you apply the front brake. When you are vertical they push the tire down on the pavement and increase friction. As you increase lean angle, your forks would push your tire less down towards the ground and increasingly out towards the side. At the extreme of the example with the bike horizontal, your forks simply push the tire sideways with no down force.

Braking while leaned puts the same deceleration stress on the front tire as it does when vertical but you don't get the same level of downward weight transfer so not as much increase in friction. I'm not sure that was any more clear. All the illustrations I found that could help are copyrighted.

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Old 05-14-2013, 01:58 AM   #125
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Originally Posted by Tripped1 View Post
Weight adds friction, which is also independent of surface area. So says Coulomb, anyway (I know there are a pile of holes in those laws) which is likely why we use the term traction for vehicles and not friction.
Tripped1 - STLE?? Do we know each other?

Friction is used to describe force resisting relative motion in static (static friction) and sliding (kinetic firction) conditions. Traction is used to describe force resisting relative motion where rolling is present. The two are linked by the slide/roll ratio. The idea behind 98% of motorcycling is to stay in traction and out of friction. In fact, the desire to remain in a tractive condition is why ABS is good.

Coulomb's mistake was to assume that observed contact area = actual contact area and neglect the impact of asperities and other roughness. As the ratio of contact area to observed contact area is relatively consistent for most situations, the reduction to observed contact area is generally sufficient. This is the first of the two significant zero errors in tribology, the second being Reynold's error.

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I think (for myself) that I often encounter folks who think that objects with more weight stop faster due to increased traction/friction. How does that calculationg work--for dummies--what does wieght increase? Simply force?
Yes and no. Weight increases the normal force, which when multiplied by the tractio coefficient (mu), also increases the friction or traction force. However, it also increases the observed contact area and the stress/strain on the two surfaces. And, there is also the inertia issue - heavier things have more intertia and will tend to stay in motion moreso than lighter things. If the entire system remains in traction and the traction coefficient is not compromised by material changes under load, the overall stopping will be similar. If the system transitions to sliding friction, forget it, that stuff is going to continue to move. So no, heavier things to not stop faster due to increased normal force, they tend to stay in motion and transition to sliding friction, which is bad.

This is why we do suspension, brake system, and tyre compound tuning. Imagine if all of it was perfect? What fun would that be?
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Old 05-14-2013, 03:52 AM   #126
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I have been following along the whole thread. I even bought the book Total control.
Thanks to everyone and thanks for being civil.

I was following the counter steering thread but......
I still gained some knowledge. Just s different bunch.

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Old 05-14-2013, 05:37 AM   #127
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Quote:
Originally Posted by Capt Crash View Post
I think (for myself) that I often encounter folks who think that objects with more weight stop faster due to increased traction/friction. How does that calculationg work--for dummies--what does wieght increase? Simply force?
Amazingly enough, I think I may be able to help with this one.

For crash investigations (no pun intended to your name), the traction or friction of the road surface is called the drag factor. Simply put, the drag factor is the mathematical equation to express how much force is required to drag an object forward in relation to the object's weight on a surface (force / weight = drag factor). For crash reconstruction, the drag factor can then be used in other equations to determine vehicle speeds (which I'm not going to go into).

Example: It takes 3,000 pounds of pulling force to drag a 4,000 pound car forward with all wheels locked. 3,000 divided by 4,000 equals a drag factor of .75. Likewise, if the same car is on the same road but the road was drenched with rain so it takes only 1,000 pounds of force to skid it forward, 1,000 divided by 4,000 equals a drag factor of .25.

So with the drag factor equation (force / weight = drag factor), you can gather that the amount of weight applied to the tire effects the traction of the road by the same proportionate amount to produce the same rate of deceleration. On the dry same road with the same .75 conditions, a 8,000 pound vehicle would take 6,000 pounds of force to skid it forward. Assuming both vehicles immediately locked up its tires to stop (to make the braking force equal), the 8,000 pound vehicle would stop in the same distance as the 4,000 pound car. This is also how a 20 pound crash reconstruction drag sled or doing a controlled test skid with a different car can accurately determine the drag factor of the road for crash investigations.

To show this to us during the crash investigation class, the instructor got a car up to a set speed and skidded to a stop. We measured the skid distance. We then loaded up the same car with the most amount of big guys as we could possibly squeeze in. We figured that it was at least an extra 1,200 pounds of weight in the car. The car was brought up to the same speed and again skidded to a stop. The stopping distance with the extra weight was very close to the same as the unloaded car. *Actually, in one test, the car with the additional weight stopped shorter but that was due to the car not having powerful enough brakes to completely lock up the wheels so it threshold braked instead of skidded.

So in theory, a 800 pound vehicle would stop in the same distance as a 400 pound vehicle if both vehicles skidded to a stop. The reality is that you cannot skid a front tire on a bike without tipping over so motorcyclists rely on threshold braking. Most 800 pound bikes are cruisers or touring bikes that have average to poor brakes so finding maximum threshold braking is more difficult than it is on a 400 pound sport bike. The 800 pound bike may not even have the mechanical braking power to threshold brake. This means the 800 pound bike won't stop as fast as the 400 pound bike because the 800 pound bike isn't using 100% of the road traction available for braking.

How this relates to cornering on a bike is left best up to the experts. My guess is that while both the 400 pound bike and the 800 pound bike have the same tire to road traction (or friction) to begin with, cornering a 800 pound bike is like stopping a 800 pound bike. There isn't as big of a gray area of the threshold zone between being upright and skidding out of control, possibly due to the additional 400 pounds wanting to continue traveling in a straight line. In addition, you also have to consider the mechanics between the bikes like the effect body positioning (would be less of an effect on the heavier bike), suspension systems and loading forces, max lean angles, frame/fork geometries, etc.


Edited to Add: I very much enjoy your videos Capt Crash. Thank you for making them.
Edited to Add 2.0: I got home from work and haven't slept yet so there may be minor inaccuracies that I might catch after I've slept. It has also been a couple years since I reviewed my text books so I might have forgotten a thing or two about the principles of it all.
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Old 05-14-2013, 05:52 AM   #128
Tripped1
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Tripped1 - STLE?? Do we know each other?
I'm working on a CE, but I had to sit through the pile physics x3, statics, dynamics, vector calc and whatnot
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Old 05-14-2013, 06:57 AM   #129
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I just luv that planted feel'in whilst drag'in the front brake into a corner.
why I hated mah Bowl Movement Werks; no feel.
Quote:
Originally Posted by Capt Crash View Post
I think (for myself) that I often encounter folks who think that objects with more weight stop faster due to increased traction/friction. How does that calculationg work--for dummies--what does wieght increase? Simply force?
Dunno 'bout all this other long talk, but bigger heavier bikes don't stop faster cuz of their extra weight, they often stop faster cuz they got real long wheel bases, 'n lower CGs, so the rear brake has mor effect. (As opposed to a sport bike's rear wheel in the air.)
Bigger bikes do have mor traction, butt they have mor weight to try to turn 'n stop.
Next time ya see folks who think bigger bikes stop (Or turn?) faster, ask 'em; why are there no 600 lb GP bikes?
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Old 05-14-2013, 07:31 AM   #130
Tripped1
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Next time ya see folks who think bigger bikes stop (Or turn?) faster, ask 'em; why are there no 600 lb GP bikes?

Indeed, that is also why there are no long or low GP bikes. (Actually the Ducati is comparatively long next to the M1 and RSV....and we all know how that is working out for them).
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Old 05-14-2013, 08:53 AM   #131
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Thanks for the answers. I had particularly forgotten inertia...
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