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Blake
| | Posted on Friday, December 27, 2002 - 02:02 am: | 
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Originally posted by CarlosT on SacBORG:
Quote:Is a stoppie the fine balance point between momentum and the gain in gravitational potential energy? Whereas
M x V = M x g x y
M = mass
V = velocity (speed)
g = gravity
y = height of center of mass (really the change in height as the bike comes up during a stoppie)
And if we were to cancel the mass out of this formula, then it would be
V = g x y
Which means that it doesn't matter how much of a fat ass you are when doing a stoppie. It's all a matter bringing the speed down to the right level before clamping the brakes.
Am I on the right track here?
Kinda, but not really.
We basically have four variables...
x, y, m, and a
where...
x = horizontal distance from center of mass of bike + rider to front axle.
y = vertical distance from ground to center of mass of bike + rider
m = combined mass of rider and bike
a = acceleration (deceleration in our case) due to braking
Be it known that weight (w) is equal to mass (m) times the acceleration due to gravity (g), a constant on planet earth for our crude purposes.
The static equilibrium equation is...
m*a*y=m*g*x
simplifying we get...
a*y=g*x
So the mass of the rider and bike have zero/nadda/nothing to do with getting the rear wheel to lift off of the ground due to braking. Of consequence are only the magnitude of the gravity (g) and the location of the center of mass vertically relative to the ground (y)and horizontally relative to the front axle (x).
That is the statics solution only.
The mass and geometry (inertia) of the bike and rider do significantly effect the rate at which they rotate once the stoppie is initiated. That would be the dynamics formulation of the problem. Al Lighton wants to work that one out for y'all too. 
Center of gravity? A silly imaginary term that. 
Enjoy some original Blake art.... ![:]](http://www.badweatherbikers.com/buell/clipart/proud.gif)
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Prof_Stack
| | Posted on Friday, December 27, 2002 - 08:12 am: |
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So is this why the salesguy cringed yesterday when, as I was about to take the XB9S for a test ride, I asked him how to do a stoppie...?  |
Newfie_Buell
| | Posted on Friday, December 27, 2002 - 08:15 am: |
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So I guess when the point of the two green arrows go past the point of the red arrows are you at the point of NO RETURN and going over the bars???? |
Blackbelt
| | Posted on Friday, December 27, 2002 - 09:07 am: |
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you guys have WAY TOO MUCH TIME ON YOUR HANDS! just enjoy doing them, instead of trying to figure them out... hahahaha |
Blake
| | Posted on Friday, December 27, 2002 - 09:24 am: |
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Newfie,
You got it. A wheelie works exactly the same way, just with the rear wheel/axle instead of the front. When the thrust at the contact patch times y exceeds the combined weight of the bike+rider times x, you are on your way to riding on one wheel.
Too much time? Nah, this is fun stuff. |
Spiderman
| | Posted on Friday, December 27, 2002 - 09:38 am: |
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NO fun stuff is out doing stoppies. Fun Physics is figuring out how much TNT and where to place it to implode a building.
Ride more, Smile more. |
Turnagain
| | Posted on Friday, December 27, 2002 - 10:19 am: |
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little math content, eh, but technique of Wheelies & Stoppies from CanadianRider.com |
Henrik
| | Posted on Friday, December 27, 2002 - 11:06 am: |
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Blake; I'm disappointed. I'd have expected at least ASCII art
Henrik |
Mikej
| | Posted on Friday, December 27, 2002 - 11:24 am: |
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__ \ | / ---
(ascii art is hard) |
Bluzm2
| | Posted on Friday, December 27, 2002 - 11:40 am: |
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Damn Blake,
Now you're scaring me again!
To quote the fly from Joe Cartoon, "Ohhhh, my freekin head" (said in your best burnt out Cheech and Chong voice).
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S320002
| | Posted on Friday, December 27, 2002 - 06:54 pm: |
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"So the mass of the rider and bike have zero/nadda/nothing to do with getting the rear wheel to lift off of the ground due to braking."
Unless of course the mass is great enough to exceed front tire traction. Sliding front tires can ruin your day.
Greg |
Ray_Maines
| | Posted on Friday, December 27, 2002 - 09:24 pm: |
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"Sliding front tires can ruin your day."
No shit! I hate it when that happens. |
Blake
| | Posted on Friday, December 27, 2002 - 10:34 pm: |
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Greg,
Traction is pretty much proportional to the weight which is directly proportional to the mass (W=m*g). At some point extreme stress on the rubber could cause it to actually fail and tear away resulting in an apparent loss of traction, but until that point is reached, and as long as the brakes are effective enough, mass will not deter a big rider from attaining stoppiness. |
Jima4media
| | Posted on Friday, December 27, 2002 - 11:39 pm: |
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But, like I said over on SacBorg, You don't have the coefficient of friction in your equation. In fact your bike doesn't have a front brake!
The coefficient of friction of the brake and the wheel will have more effect than the mass of the bike and the rider.
Simple physics indeed. |
Blake
| | Posted on Saturday, December 28, 2002 - 12:32 am: |
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Oh cripes, no front brake! I guess we should next investigate the physics of a collision? LOL! Yeah, gee, I left out the brake lever, carburetor, and fuel and all the equations of thermodynamics governing the operation of the engine too. Silly me.
But seriously Jim, as discussed, the bike and rider are a system in equilibrium, the brakes are internal to that system; their coefficient of friction simply relates to how hard the rider must squeeze the brake lever to achieve the requisite deceleration (a). The coefficient of friction of the tire/pavement of course must be sufficient to withstand the decelerative force. I should have stated those assumptions. But hey, we are talking about sport bikes here right?
So you see, there is no use for the coefficient of friction in the basic equation of static equilibrium for the system of interest, the bike and rider. How the bike/rider get there is inconsequential to the discusion, where the overall relationship of the basic forces of equilibrium is inherently germain to the topic.
Now, if you want to know how hard you must squeeze the brake lever to initiate a stoppie, yeah, then we would need to know the coefficient of friction for the brakes. We'd also need to know the mechanical advantage of the brake lever, and the ratio of master cylinder piston bore area to total caliper piston bore area. We'd also need to know the ratio of the effective brake disk radius to the tire's outer (contact patch) radius.
Here's an exercise for you that might help illustrate the concept of "system equilibrium". Sounds easy. Try to lay out the static free body diagram (a diagram like mine above) for only the front wheel assembly. That is the wheel, axle, tire, and brake disk assembly... show all the forces acting upon that system as a whole during hard braking. Remember they must all sum to zero along with any moments that might result.
Some hints... the forks impart loads to the axle, the brake pads impart load to the brake disk, and the pavement imparts loads to the tire.
The loading consists of... dead weight, braking force, inertial loads. It's *just* simple physics. |
Jima4media
| | Posted on Saturday, December 28, 2002 - 03:04 am: |
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Here is another exercise. Which will stop faster, a Formula 1 car with 4 carbon brakes, or a Superbike with two brakes? And Why?
Neatness counts. |
Dynarider
| | Posted on Saturday, December 28, 2002 - 07:31 am: |
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The Formula 1 car will stop harder, accelerate harder & corner faster. The only reason I can think of why it would stop faster is increased brake pad surface area & a much larger contact patch of rubber on the ground. |
Hootowl
| | Posted on Saturday, December 28, 2002 - 11:58 am: |
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Jim, cold brakes or hot brakes? |
Jima4media
| | Posted on Saturday, December 28, 2002 - 12:43 pm: |
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Hoot,
Hot or cold. They will be hot after the first corner. Formula 1 brakes glow red hot at over 1,000 degrees.
The Formula 1 car will stop quicker than a Superbike because of the braking surface compared to the weight of the vehicle, the width and stickyness of the tires, it's center of gravity, and the downward force of body surfaces, and the fact the the driver is strapped into the seat and can withstand the 4G deceleration compared to a 1G deceleration of a Superbike.
There is so much aerodynamic down force on a F1 car that they could be driven upside down on a road and stick to it. That down force keeps the tires from locking up and skidding at faster speeds.
Interesting stuff.
Jim
X-2.5 |
Ray_Maines
| | Posted on Saturday, December 28, 2002 - 05:27 pm: |
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Blake
| | Posted on Sunday, December 29, 2002 - 12:33 am: |
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Jim,
Factors affecting superior braking of an F1 car versus a superbike?
Braking surface/weight? Nope.
Width and stickyness of tires? Nope.
Lower center of mass? Yep.
Downforce? Yep.
Driver restraints? Yep.
The lower center of mass is the main reason. The F1 car will not be limited by a tendency to endo upon extreme braking like a superbike.
Where do you get a limit of 1g for deceleration of a superbike?  |
Hootowl
| | Posted on Sunday, December 29, 2002 - 12:42 am: |
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Jim,
That's why I asked. Those carbon brakes work better the hotter they get. |
Jima4media
| | Posted on Sunday, December 29, 2002 - 04:24 pm: |
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Blake,
So you think you can put motorcycle tires on a F1 car, two drum brakes from the 1930's, and race in a car that weighs two tons, and it is not going to be any different in performance that Michael Schumacher's Ferrari?
Ferrari should be made aware of these facts of yours, because they are wasting a lot of money on tires and brakes!
;-) |
S320002
| | Posted on Sunday, December 29, 2002 - 07:32 pm: |
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"Traction is pretty much proportional to the weight which is directly proportional to the mass (W=m*g)."
According to your simplistic explanation of traction stoppies should be just as easy to do on a Bagger as they are on an XB. Also the Top Fuel Nostalgic dragster that I sometimes pit for would never be able to accelerate at 4 Gs. It doesn't have a rear spoiler so it doesn't generate 4 Gs of downforce.
If I had hours of time to spend I could help you gain a better understanding of the physics of traction.
Your explanations entirely ignore the dynamics of physics. Not very realistic in the real world.
Greg |
Blake
| | Posted on Monday, December 30, 2002 - 12:20 am: |
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"So you think you can put motorcycle tires on a F1 car, two drum brakes from the 1930's, and race in a car that weighs two tons, and it is not going to be any different in performance that Michael Schumacher's Ferrari?"
Uh, No. I was addressing your original question...
"Which will stop faster, a Formula 1 car with 4 carbon brakes, or a Superbike with two brakes? And Why?"
In addressing that question, I am saying that the F1 car will stop more quickly and that the factors affecting how much more quickly an F1 car can stop versus a Superbike are primarily...
1. The F1 car's lower center of mass (keeps it from doing an endo).
2. The F1 car's aerodynamics (enhances downforce and thus braking).
3. The driver restraint system (for obvious reasons).
Your original queery mentioned nothing of "racing performance" or altering the performance an F1 car by swapping parts with a motorcycle, rather you only asked which would stop more quickly, and why.
Superbikes use drum brakes from the 1930's?
A Formula One car weighs two tons?!? I think they weigh, with rider, fuel, coolant, and lubricant, significantly less than half that, somewhere around 1,500 LBs (3/4 of a ton).
Finally, when you offer that an F1 car would stop better with its own tires and brakes versus those of a motorcycle you have completely changed the crux of the debate. Obviously that scenario is a bit different than stating that an F1 car stops better than a superbike. In either case of course, I would agree with your conclusion.
Mate a big fat formula one wheel/tire and F1 brake on the front of a motorcycle, and it will stop no better than with its own brakes and tire. Why? The three reasons given above. How's that for throwing your own misdirection back at you? heheheh
And as to the original question (Would an F1 car stop faster than a superbike and why) I'm simply saying that the significant factors responsible for that do not include differences in tire or brake technology, but primarily the lower center of mass (F1 car will not endo), aerodynamic downforce (permits harder braking), and driver restraint system (keeps rider in vehicle), not by virtue of more/bigger tires and more/better brakes.
You may want to consider... Is an F1 car/driver that weighs ~1,500 LB with four brake disks better equipped to stop quickly than a 550 LB superbike/rider with two brake disks on the front wheel? Hmmm, 1,500LB/4=375LB/brake-disk for the F1 car where the superbike has 550LB/2=275LB/brake-disk. So then, which brakes have less work to do? Also, what diameter are the F1 brake disks versus those on the Superbike? Are larger brake disks more effective? And... don't they use carbon disks on superbikes too?
Still haven't said how you end up with a 1 g deceleration limit for a superbike. I wouldn't be surprised that at very high speeds a superbike can decelerate at near 2g's. I don't think the rider's full weight comes to bear on his arms. I think his legs squeezing the specially shaped tank might also help carry some of that load.
Give up on solving the free body diagram for the motorcycle front wheel/brake/tire assy? |
Blake
| | Posted on Monday, December 30, 2002 - 01:05 am: |
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"According to your simplistic explanation of traction stoppies should be just as easy to do on a Bagger as they are on an XB."
I don't know how you make that fantastic leap in logic. I also didn't know I was trying to explain traction. I thought I was answering Jim's questions. Maybe you should reread the thread.
Regardless, baggers don't usually come fitted with sticky type sporting tires let alone a tire comparable to a superbike racing slick, do they? A bagger/rider has a very low center of mass compared to a superbike/rider, right? Baggers generally have weak front brakes where superbikes have extremely strong front brakes, right? Maybe you should reread my comments and or the question originally posed by Jim?
"Also the Top Fuel Nostalgic dragster that I sometimes pit for would never be able to accelerate at 4 Gs. It doesn't have a rear spoiler so it doesn't generate 4 Gs of downforce."
You imply that you would need 4 g's worth (4 times the weight of the car/rider) of downforce to support 4 g's of acceleration. Nope, not even close.
Do you have any idea of the effective coefficient of friction between a top fuel dragster's rear tires and the sticky ole drag strip? Hint... it is WAY greater than 1.0. Even an all out top fueler with airfoil/wing has little to no aerodynamic downforce until speeds climb towards triple digits. How then do the top fuelers accelerate so fiercely right off the line? Answer... traction (read, very high effective coeficient of friction).
"If I had hours of time to spend I could help you gain a better understanding of the physics of traction."
Call me collect, I'd love to be educated on the subject (number is in the RAN list).
"Your explanations entirely ignore the dynamics of physics. Not very realistic in the real world."
The "dynamics of physics" eh? Well, I think I know what you are trying to say. What should be clarified is that I was only intent on answering Jim's question, not writing a technical treatise on traction. Again, as far as I know, the racing slicks used on superbikes are comparable in their tactile properties as the racing tires used on Formula One cars. In fact, F1 cars no longer even use slicks, do they?
Plain ole DOT treaded sporting street tires are capable of supporting 1g+ loading. Racing slicks?
Finally, yes I do understand very well that the interaction betwen tire and tarmac is much more complex than a simplistic "coefficient of friction" would lead one to believe. |
Sportsman
| | Posted on Monday, December 30, 2002 - 01:20 am: |
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You guys are full of physics. Use a set of EBC pads for a EBC rotor, but use the stock rotor. You will do a stoppie, ruin your rotor, but you will do a stoppie. |
Blake
| | Posted on Monday, December 30, 2002 - 01:39 am: |
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With a superbike racer sitting up and hard on the brakes at 180 mph, what do you imagine the total aerodynamic drag acting on bike/rider might be?
Answer... as much as or more than 300 pounds.
Does that count towards effective braking force?
Answer... heck yes. |
Jima4media
| | Posted on Monday, December 30, 2002 - 02:51 am: |
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Blake,
I was just responding to your statement ----
"Factors affecting superior braking of an F1 car versus a superbike?
Braking surface/weight? Nope.
Width and stickyness of tires? Nope. "
All I'm saying is the braking surface of 4 hot carbon fiber Brembo brakes on a light 650 Kg car are able to out-brake a Superbike with a pair of Brembo carbon fiber brakes (Moto GP only - They are outlawed in WSBK and AMA Superbike)
And those fat sticky Bridgestone tires also figure into the equation. You could put them on a Superbike, but then you wouldn't be able to turn it. ;-)
As far as sitting up at 180MPH on a Superbike - Do you think that is possible? Ask Richard Nallin if he as ever sat up at 180, and lived to tell about it.
And any more than about 1.5 g of deceleraton, and you will not be able to stay on the seat of a bike, even with your knees velcro'd to the tank. ;-)
Jim |
Dynarider
| | Posted on Monday, December 30, 2002 - 07:32 am: |
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Dont let Blake bullshit you with a bunch of math equasions. Im not a math or physics major so for all he know he is totally full of shit on this subject. Course he could be right, but I prefer to think he's full of shit |
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