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Inertia Moment for Racing Motorcycle Wheels 
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#1
Mar1308, 04:40 PM

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Hi, I'm hoping you kind folks can help me out. I'm a motorcycle racer, and I'm trying to find a *simple* formula for calculating how much mass addition/reduction to a wheel will increase/decrease the force required to change the direction of the motorcycle when at speed.
Basically, to turn a motorcycle, you push on the handlebar in the direction that you want to turn, which "deflects" the wheel. The problem is that when you're going fast, the wheels (and other engine internals) are moving fast and so steering the bike at speed a certain amount requires more force than steering it the same amount at a lesser speed. So I'm interested in knowing how to calculate, for a (for my purpose) uniform disc rotating around its center axis, how much difference additional or subtractional weight would make on the force required to turn the bike at a given speed. For example, if the wheel weighs, let's say, 12 pounds, I want to know how to calculate how much "force" is required to pivot it on its axis at, say, 100mph. I need a formula that I can understand, because I've been out of college physics and calculus so long that I don't remember what all of the greek symbols represent, and so on. Thanks in advance for any help with this problem. 


#2
Mar1308, 05:20 PM

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#3
Mar1308, 06:05 PM

P: 11

At speed, a motorcycle steers by leaning over in the direction of the turn, and that lean is initiated in a normal circumstance by pushing mostly *downwards* on the handlebar that's on the same side as the desired turn, and by using other parts of the body to create the forces necessary to initiate the turn. In fact, racing bikes have steering stabilizers to prevent the handlebars from actually moving, because all of the steering is done by leaning, and any excessive movement of the handlebars could upset the suspension, or potentially succumb to a harmonic called "headshake" or a "tankslapper" where the handlebar violently moves backandforth out of the rider's control. 


#4
Mar1308, 06:19 PM

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Inertia Moment for Racing Motorcycle Wheels
The other main technique for steering a streetbike at speed (mainly lighter sportbikes), is "body steering". That's where you use weight shifts on the footpegs to turn the bike, and let the front end find its own turn angle that matches the lean angle and bike speed. Racers typically use a mix of these two techniques, according to personal preference. Personally, I prefer 100% body steering, whether on the racetrack or on the street, but again, that's personal preference. So to your question, kl3640.... The two basic concepts you will use to do your calculations are "Moment of Inertia" and "Torque/Precession". The moment of inertia ratios with the mass, and with the distance the mass is out from the axle: http://en.wikipedia.org/wiki/Moment_of_inertia If you click on the Table of Moments of Inertia in that link, you'll start to see some basic formulas for the I values of various geometries. A real sportbike wheel is a mix of some mass between the axle and the rim (to support the rim), and mass at the radius of the rim (and the tire). You can actually measure the I for a real wheel by spinning it up with a known torque (get out your torque wrench and a stopwatch, and get creative). Once you know the moment of inertia I for your two (different) wheels, you will use the equations related to precession to figure out how much torque it takes to displace the wheels and lean the bike. I honestly am no expert on this part, but I think the correct term to look into is "torqueinduced precession", where the spinning of the gyroscopic I resists the motion that you are trying to cause with the torque: http://en.wikipedia.org/wiki/Precession#Torqueinduced For your countersteering case with no body steering input, your torque is put into yawing the front wheel, to get it out from under the CG of the bike, and cause tipin. For the bodysteering example, the torque is from the unbalanced weight on the two footpegs, resisted by the gyroscopic spinning of the two wheels. I'll have to look into this a bit more to give a better answer. Let us know how far you get with this intro, though. 


#5
Mar1308, 06:35 PM

P: 11

Berkeman, that's what I was hoping for  thank you. I'll proceed with this and let you know how it works for me (I'm ultimately hoping to get a simplified equation that somehow represents the different in work required to initiate a turn, so that different weights of a wheel, i.e., before and after powdercoating and the related weight difference, can be used to determine how they will affect handling performance particularly related to steerin and transition).
Also, a related question if you don't mind: assuming that nothing changes about the wheel except for weight (i.e., the diameter and other relevant realworld dimensions remain constant), can I just ignore the tire as it would be a constant in both circumstances? 


#6
Mar1308, 06:41 PM

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To induce a lean on a bicycle, motorcycle, or unicycle, a rider steers outwards to initiate a lean. The outwards steering can be the result of a rider applying an outwards steering torque force on the handlebars (countersteering), or the result of the rider leaning inwards, causing the motorcycle to lean outwards (since the center of mass won't move sideways without a sideways force at the contact point between tires and pavement (or a side wind)), and the selfcorrecting geometry causes an outwards steering reaction to the motorcycle being leaned outwards. Deliberate countersteering is best, since body leaning doesn't provide as quick as a response, and at high speeds on a motorcycle, body leaning provides no lean inducing steering response.
The rate of precession from gyroscopic effects is less than the desired rate of lean (roll) for a normal situation. The faster the speed, the more pronounced this effect becomes. At moderately slow speeds, a motorcycle has a tendency to remain or return to a vertical state, while at high speeds a motorcycle has a tendency to remain at it's current lean angle. At high speeds, it takes about the same amount of countersteering effort to decrease lean as it does to increase lean. From a previous post: Two wheeled vehicle vertical stability is due to steering geometry, specifically trail. There are radio controlled motorcycles that don't require any special algorithims or gyro sensors to work, they just use a lot of trail, and a steering servo with a low resistance to small "noncommanded" movements caused by trail effect, allowing the self correction process to work. Trail is the distance from where the front tire pivot point axis would intercept the pavement, back to the point where the front tire actually makes contact with the pavement. When a bicycle is vertical, the trail creates a castor effect (like the wheels on a shopping cart), creating a tendency for the front tire to move in the direction traveled. When a bicycle is leaned, the trail creates a inwards steering torque force on the front tire. This is because contact patch moves laterally when steering and vice versa. Suspend a bicycle off the ground. Steer left, and the contact patch area on the front tire will move right and vice versa. This lateral motion is relativly large on a radio control motorcycle, and moderate for human controlled bicycles / motorcycles. To see this effect, a person can hold a bicycle by the rear seat and lean it over, and the front tire will steer inwards. Getting back to a leaned bicycle, gravity results in downwards force on the center of mass, and the pavement in turn generates an upwards force on the tires. At the front tire of a bicycle, this upwards force is applied "behind" the pivot axis and causes the front tire to steer inwards, and given suffiecient inwards steering and speed, the lean angle of the bicyle will be reduced until it is vertical again (it may overcorrect due to momentum). The amount of trail effect determines the minimal speed required for vertical stability. At or above this minimum speed, a bicycle will be vertically stable, autocorrecting for any reasonable amount of lean introduced. If there is excessive trail, flex, or momentum, in the bicycle, constant overcorrection can occur, resutling in speed wobble. Gyroscopic reaction generates lean angle stability (as opposed to vertical stability). As speed increases, a bicycle will tend to hold a lean angle and resist changes in lean angle, including the vertical stablity reaction from trail. In the case of motorcycles at high speeds, 100+mph, the lean stability dominates, and the motorcycle just holds a lean angle with no perceptible tendency to straighten up. 


#7
Mar1308, 06:42 PM

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So the contribution from the tire will be constant, and the contribution from changing the wheel shape or weight distribution would be slightly nonlinear with weight, depending on the distribution of that weight with respect to distance from the axle. I doubt powder coating will make much difference  what is the change in overall weight for the full wheel+tire assembly for powder coated versus nonpowder coated? What is it for the front and for the back wheels? 


#9
Mar1308, 11:36 PM

P: 11

I argue that because on the track while most of us do what is called "countersteering" (i.e., pushing the bar in the direction of the intended turn to initate the lean), we also set up our bodies to shift the center of gravity of the bike for the turn (i.e., hanging off the side and dragging our knees) to enter the corner faster and carry more speed through the turn. While that itself isn't for the purpose of intiating the turn, what we do in the process is shift weight to the appropriate foot, release pressure with the inside thigh and exert pressure with the outside thigh. Then, because we also have shifted our bodyweight towards the inside of the turn, it helps to affect, in conjunction with the steering inputs, the lean of the bike in the direction of the intended turn. Depending on what is considered "high speed," I can weave my race bike back and forth, and around a complete turn in fact, with nothing but my legs, with my hands completely off the bars. I'm not sure why the move was ever dubbed "countersteering" because the goal isn't to really turn the handlebars in the classic sense, but to rather lean the bike to the side of the intended turn. In that respect, pushing on the bars in that direction (and actually, as we do on the track, pulling on the opposite side) makes more intuitive sense to me. But I suppose that the move does technically deflect the wheel in the opposite direction, for however miniscule of a degree and moment. 


#10
Mar1308, 11:45 PM

P: 11

The reason that I want to be able to do the calculation is because I want to figure out things like the disadvantage of powdercoating in quantitiave terms, but also for things like ceramic vs. steel bearings, etc. Also, I want to be able to sniffout if things like this are BS or not: "Each ounce of weight reduction on the rim is equal to about 24 pounds of weight at 100MPH!" That's from the website of Mavic, a manufacturer of highend racing wheels for motorcycles & mountainbikes. 


#11
Mar1408, 01:07 AM

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Imagine a linear forwards force applied to the axis of a hollow cylinder (a very thin wheel). This forwards linear force is opposed by the friction force which peforms the angular acceleration (rotation) of the hollow cylinder. The angular inertia of a hollow cylinder, Ic = M x R^2 (M = mass of cylidner, R = radius). In equation form: LF  FF = M LA, LF is linear force, FF is friction force, M is mass, and LA is linear acceleration. LF is a constant. AA (angular acceleration) = T (torque) / Ic AA = LA (linear acceleration) / R (assuming no slippage here) LA = AA x R Torque = FF x R LA = ((FF x R) x R) / (M x R^2) LA = FF / M FF = LA x M So substituting in the first equation: LF  FF = LA x M LF  (LA x M) = LA x M LF = 2 x (LA x M) LF / M = 2 x LA LA = 1/2 (LF / M) So linear acceleration of a wheel is 1/2 of what it would be if it wasn't for the increase in angular speed. For example, if friction force FF=0, then you'd have LA  0 = LF/M. So weight at the perimeter of the tire affects acceleration by twice as much as the same weight would on a nonmoving part of a motorcycle. 


#12
Mar1408, 11:28 AM

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When you use a combination of body steering and countersteering, you are using the yaw pushpull on the bars to help displace the CG of the bike, to help the bike lower itself into the lean. In my experience (on the track), countersteering alone is just not enough to control a bike at speed through complex curves. I tried and tried to hit the Corkscrew right with just countersteering (before I learned bodysteering), and my reactions were just not fast enough to control the complex line. But with bodysteering, controlling the path of the bike becomes much more intuitive, since the bike is steering the front end itself, and you are only concerned with shifting your weight to do the overall line control. The main reason that I switched to 100% body steering for my street riding, is that countersteering relies on you making these nonintuitive pushpull motions with your arms, and when you get into emergency situations (cars dodging in front of you, or you getting into a decreasingradius turn too hot), the first thing that happens is that your arms tense up, which decreases your ability to deal with the emergency. Not good. Using 100% body steering avoids this handicap, in my experience, and makes emergency maneuvers much more of an intuitive deal. EDIT  Here's the Corkscrew at Laguna Seca, for those who don't know the turn already: http://www.corgifan.com/blogger/americanflyers2.jpg http://www.motorcycleusa.com/photos/corkscrew.jpg 


#13
Mar1408, 01:25 PM

P: 11

Separately, thanks for this info, this is very helpful to me. Perhaps you can explain something to me, to make it easier for me to calculate the difference in modifications to the static mass of an object that rotates in the direction of travel and the work required to overcome it at speed (i.e., the amount of effort I need to exert to divert a bike at speed from a straight line in order to initiate a turn): Is there a proportional relationship between the static mass of an object, such as a wheel, and the rotational inertia, such that if I increase the static mass of the wheel uniformly (i.e., NOT just towards the center or the edge) by some percent X, I can then derive the increase in the rotational inertia by that same percent X? To be honest, I'm not really interested in the absolute #'s, I'm really just interested in figuring out how I can simply calculate the effect of a weight increase/reduction on a rotating object (rotating in the direction of travel) on the effort required to deviate the vehicle from its current course (i.e., initiate a lean). Is there a simple proportional relationship that I can use to infer how much more or less work I'll have to do steer the bike? Even if it's not a lot of work, it makes a difference because there are some looong races (like the last one was 52 laps at Daytona) so that little bit adds up over time with respect to muscle fatigue. 


#14
Mar1408, 06:26 PM

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http://www.vf750fd.com/blurbs/countercode.html Note the "no bs" motorcyle is a 600 road racer replica, with little trail effect. When there is little trail effect, then body leaning can't be used to steer a motorcycle. On street motorcycles, at least ones with a lot of trail effect, I can sit up with my hands off the handlebars and do mild turns with body leaning, but this only works for a limited range of speeds, about 35mph to 45mph, depending on the motorcycle. As mentioned before at racing speeds, 100mph, the trail effect is virtually gone, and body leaning will do virtually nothing. I've read a few comments about this with regard to Daytona, where the riders go onto the banked section, exiting the banked turn at around 180mph, and nearly horizontal. It takes a huge amount of countersteering effort just to get the bikes back to vertical, one of the comments that stuck in my mind is that a rider either knows about countersteering when coming out of the banked turn, or the rider ends up in the infield (plowing into something hard at high speed is not good). Thin circular hoop: I = M x R^2 Solid disk: I = 1/2 x M x R^2 Say you only increase the weight of the spokes of a wheel but not the rim, and treat the spokes of a wheel as a solid disk, using a front wheel in this example. wheel radius = 8.5 inch (17 inch diameter wheel) tire radius 120x70x17 = 120mm x .70 + 8.5 in = 3.3 in + 8.5 in = 11.8 in wheel radius = .72 of total radius Repeating the math as before: Iw = angular inertial of wheel, treated as a solid disk: Iw = 1/2 M (.72 R^2) = .26 M R^2 LF  FF = M LA AA (angular acceleration) = T (torque) / Iw AA = LA (linear acceleration) / R (assuming no slippage here) LA = AA x R (left this out before) Torque = FF x R LA = (FF x R^2) / (.26 M x R^2) LA = FF / (.26 x M) FF = LA x .26 x M So substituting in the first equation: LF  FF = LA x M LF  (LA x .26 M) = LA x M LF = 1.26 x (LA x M) LA = LF / (1.26 M) So any weight added to a solid disk would have 1.26 times the effect. 1 lb added to to the spokes would be the same as 1.26 lbs added to a nonmoving part of the motorcycle. For the case of adding weight to the rim of the wheel: Iw = angular inertial of rim, treated as a thin hoop Iw = M (.72 R^2) = .52 M R^2 LF  FF = M LA AA (angular acceleration) = T (torque) / Iw AA = LA (linear acceleration) / R (assuming no slippage here) LA = AA x R (left this out before) Torque = FF x R LA = (FF x R^2) / (.52 M x R^2) LA = FF / (.52 x M) FF = LA x .52 x M So substituting in the first equation: LF  FF = LA x M LF  (LA x .52 M) = LA x M LF = 1.52 x (LA x M) LA = LF / (1.52 M) So any weight added to a hoop with .72R radius would have 1.52 times the effect. 1 lb added to the rim would be the same as 1.52 lbs added to a nonmoving part of the motorcycle. Basically the effect will be 1.00 + the coefficient for angular inertia. Actual radius, doesn't matter, just the relative radius of the spokes, or wheel to the outer radius of the tire (note, that the effective radius is a bit smaller, because the contact patch deforms, but I don't think you need numbers that exact). In your case, weight added to the spokes has a 1.26 multiplier, and weight added to the rim has a 1.52 multiplier. For the back wheel, the effect is less, because the tire is larger, for a 190/50/17, its 3.75 inches. This reduces wheel radius factor to .69, and weight added to spokes would have a 1.24 multiplier and weight added to rim would have a 1.48 multiplier. To caculate tire height from a tire size, width / profile / diameter width is given in mm, profile is a percentage, and diameter doesn't matter Tire height in inches = width x profile / 2540 Total diameter = diameter + 2 x tire height in inches. A simpler solution is to not add weight to the wheels, as I don't seem the value of cosmetic additions to a racing motorcycle. Another soluton would be to work out so fatigue isn't a factor on the longer races. 


#15
Mar1508, 01:33 PM

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#16
Mar1508, 10:28 PM

P: 11

When we try to adjust geometry on our bikes we usually do it be raising/lowering the forks within the triple clamps and adjusting the shock to change the rear ride height. The specific wheels (17" vs. 16.5") and types of tires (sidewall height, rim width, and tire width) are also contributing factors. The problem isn't the addition of weight to wheels, the problem is where to make improvements on a bike given a limited budget. Powdercoating wheels is one example, but a better example would be: I have a limited budget of say $2000 for improvements to my race bike. Would I be better off spending all of it on Marchesini alloy wheels which reduce weight from stock by X%, or would I be better off spending it on a combination of other things, such as Al subframe, Ti bolt kit, etc. So then my question is, what is the easiest way for me to tell how much more a weight reduction/increase on a rotating mass affects the effort of riding the bike at speed versus the same reduction/increase in static sprung or unsprung mass? Thanks. 


#17
Mar1508, 10:36 PM

P: 11

At speed, let's say coming off a straightaway for a left hand turn, I would first slow down using a combination of the brakes, engine braking (also to get the engine in to the right transmission gear to use through the turn, when I get back on the throttle), wind resistance (sitting up and putting my legs out to "catch the wind"). Then, as I approach the turn, but before I enter it, I would "hang" off the left side of the bike. What that means is that with my hands on the handle bars I raise myself up off the seat of the bike by using my legs and the balls of my feet on the footpegs, and then move my buttocks off of the seat in the direction of the intended turn. At this point approximately half of my buttocks are on the seat, and half are hanging off in the direction of the intended turn. My right thigh is up against the right side of the fuel tank, and my left leg is pivoted away from my body with the knee pointed outwards and down. My right arm is extended almost straight across the top/right side of the tank, while my left is bent at the elbow and out a bit. So what I was saying is that I think you meant that in order to manage the direction of the bike while I am momentarily in this position but before I want the bike to turn aggressively, I exert varying amounts of pressure on the "inside" handlebar (the handlebar on the side of the bike the direction of the turn) and "outside" handlebar (the handlebar on the side of the bike opposite of the turn). I modulate that pressure depending on how much I want the bike to in the direction to which I'm hanging off. Is that pretty much what you had meant? Again, my apologies for any confusion that I may have caused. 


#18
Mar1508, 11:31 PM

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However, it would seem that body shifting would consume much more energy than any countersteering effort, so I'm not so sure why you are focused on the counter steering effort. Regarding body shifting, I've seen some extremes where the riders outside thigh is the only thing touching the seat, and the rider's outside foot is just dangling, not even on the peg. This was in an older video (back in the Freddie Spencer / Kevin Schwanz era), so I'm not sure that riders still do this. Lighter wheels and tires always help, but your probably stuck with the tires, and kinetic energy and gyroscopic effects don't seem like they would be as important as the sprung to unsprung ratio benefit of lighter wheels and/or brakes. However, if you're not racing on rough tracks, then the sprung to unsprung ratio could be another case of diminishing returns. I would recommend asking questions like this at a motorcycle racing oriented forum. The people here at this forum can answer techincal stuff, but what you need is good advice from fellow motorcycle racers, or better yet, the guys who design racing motorcycles. 


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