# Lean Angle == Turn Radius?

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OK...

Tweek, David, et al...

I lack the skill to create a good diagram. If anyone knows some place to copy/paste vector arrows and such, please let me know.

Here are the numbers:

Given:

mass = m = 600 lbs = 272.72 kg

radius = r = 50 ft = 15.38 m

lean angle ( theta θ ) = 50°

Find:

velocity (v) = ?

Formulae:

gravity (g) = 9.8 m/s2

Fgrav = Fvert = m • g = 272.72 kg • 9.8 m/s2 = 2672 Newtons

Fnet = Fhoriz = m • a

And: circular acceleration (a) = v2/r

Hence, substituting for (a): Fhoriz = m • v2/r

Being that we know θ = 50° and Fvert = 2672 N, we can use our trigonometry to find Fhoriz using theta's tangent and substitute the value of Fhoriz in the equation above leaving v as the only variable to solve:

TAN 50° = opposite/adjacent = Fhoriz/Fvert

1.191 = Fhoriz/2672 N

Fhoriz = 1.191(2672) = 3183.2 N

Substituting: Fhoriz = m • v2/r

We get: 3183.2 N = 272.72 kg • v2/15.38 m

v2 = (3183.2/272.72) 15.38 = 179.516

v = 13.398 m/s = 48.234 kph = 29.9 mph

racer

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Any comments or questions? Tweek? David? Anyone?

I realize this site is dedicated to riding skills, not math; but, the physical concepts behind the math are fundamental to the riding skills. The "why' behind the "what", so to speak. One can follow a recipe for what to do in a simplified single turn, but, I think that understanding the principles, learning how to apply the fundamentals leads to the ability to adapt to different situations and requires at least some understanding of the "why" ...

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Looks right to me. The cornering angle numbers they show on MotoGP are probable bike angle instead of center of gravity/contact patch angle. We're going out of town for several days. When we return I'll take a stab at estimating the difference although I'm sure it's been done already.

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Thanks.

Have a safe trip.

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Sorry about that - work intruded.

I need to spend some quality time with my white board to grok what was done. But with the quick glance I gave it - Should have been able to get that myself. nothing more than highschool physics. Thank you.

in general I don't think a super accurate model is needed to get to what I was after. simply - about 30mph is how fast you need to go in order to lean the bike over 50degrees.

I've gone back and reread the relevant bits on TOW2. So the answer to my original question - lean angle == turn radius is false. the two variables are part of a larger system affected by velocity and weight. For instance - keeping the lean angle constant, and increase velocity. Radius increases. oh well.

For practical applicaitons (we don't ride on skid pads - at least not until next year when DMG takes over AMA) we also have to consider how quickly you turn the bike. Keith goes on at length about this. Basically, the fastery you bring the bike to max lean angle the less lean angle you need. Which seems like fun feedback loop to have to deal with. IE - first pass through I lean the bike over slowly, scape my knee through the corner, finish off my latte and exit the turn. Next pass I'm in a hurry, using the same turn in point I slam the bike over like I'm Rossi and end up looking at at surprised corner worker so I have to either roll on more throttle or stand the bike back up. Third time I keep the same TP and still slam the bike over but this time I don't gone over as much. Next time same TP, same steering rate but more velocity. Knee scrapes, no surprised corner worker.

anyway - I think this horse is dead. I get it and have the math in hand so I can think about it further. My big take away from this though is how critical the steering rate is in the real world. Next week we'll work on that.

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in general I don't think a super accurate model is needed to get to what I was after.

Nnot sure what you mean by "super" accurate, but... the number significant digits after the decimal point in the solution are required to solve the math accurately. Remember how far off the reverse calculation was when I rounded off just a few hundredths?

simply - about 30mph is how fast you need to go in order to lean the bike over 50degrees.

I don't mean to be pedantic, but... about 30 mph is how fast you need to go to lean the bike over 50 degrees for the given turn radius (50 ft). I'm sure that is what you meant. I just want to be clear for those who might not have your level of understanding.

So the answer to my original question - lean angle == turn radius is false. the two variables are part of a larger system affected by velocity and weight. For instance - keeping the lean angle constant, and increase velocity. Radius increases.

Why is that? Why does the radius increase with more velocity? I mean, sure, there is more horizontal "cornering" force working to push the bike wider in the turn, but, mechanically speaking, what happens on the bike to cause or allow that to happen? The radius of the wheels doesn't change, neither does the wheelbase really. Are the tires sliding or "creeping" sideways? Does more weight shifted to the rear cause the front end to lift and alter the rake and front wheel angle into the turn? Or lengthen the wheelbase enough to describe a longer arc at the same lean angle? But, wait... isn't the rear wheel what dictates the direction of a motorcycle once it is leaned over?

The football half-back's radius is dictated by where he puts his foot which leads to how far he needs to lean to remain balanced. The side cut of the snow skier's ski largely dictates the radius the ski will carve at a given lean angle. And the same is true for the "rocker" or curve in the blade of an ice skate. These examples do not seem to follow the same formula as a motorcycle in a turn in that their turning radius is largely mechanically "pre-set".

So, what gives?

???

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One thing to keep in mind is the the weight of the rider and bike do not matter, in principle. An easier way to compute the above is:

Given velocity v in m/s, radius r in m, using g (gravitational acceleration of 9.8 m/s2),

lean angle = invtan(v*v/r/g) (lean angle from upright, so upright=0 deg.)

If you have the lean angle theta and want to find v, do this:

v = squareroot(g*r*tan theta)

Keep in mind the angle is measured to the center of mass of the bike and rider as a whole, so if you are leaned off with your body the bike (and therefore tires) can be more upright.

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Hi Stevo,

Yep. Given the known constant acceleration of gravity and the radius, circular acceleration can be found using either velocity or theta without needing to know the mass. And, whether a rider weighs 100 lbs or 200 lbs, the bike will (theoretically) lean the same amount for a given radius and velocity (assuming the rider is not hanging off). Thanks for pointing that out.

However, that said, I think that simply plugging in that mathematical 'shortcut' sorta sidesteps the fundamental physical concepts involved (like component force vectors, etc.) that many foundational riding techniques and concepts are based on. For instance... hanging off. And why hanging off allows you to corner faster.

In other words, while it is true that the ratio of the component force vectors (as opposed to the actual forces) is what is critical to mathematically solve this particular problem, I think seeing the actual individual forces drawn out in a free body diagram might be critical to learning/understanding why.

racer

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[All of my 'advanced' math is linear algebra and combinatorics - I can count bottle caps. Some how I missed out on all the good calculus classes, only got 1 & 2 - so bare with me]

I'm not 100% sure why the radius increases w/ more velocity. However, I'm pretty sure more velocity (keep all other variables constant) won't decrease the radius. I'm guessing the radius increases b/c the lateral force exceeds the rear tire's ability to grip. However, that would mean that as the radius increased the lateral force decreased. Does the forumla support that?

Factoring in things like friction though starts making the model more complicated than necassary. But I guess it's sort of a logic thing - if velocity has no baring on the turn radius why to race bikes have such big brakes? If velocity isn't a factor then Nicky Hayden would keep the same velocity through every turn and hold the bike on its red-line through out the whole race. But that isn't what happens. Or am I making a hash of things?

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No, I don't think you are making a hash of things.

I agree that adding in things like sliding and friction will unnecessarily complicate our admittedly simplified model, but, I don't believe the model is over-simplified for this particular question either, ie. I don't think we need calculus to answer the question, at least not conceptually.

That said, if the rear tire is sliding, won't the lean angle tend to decrease (er...lean more from vertical)? I am not sure what the answer is. My gut says that it must be mechanical. It says the front wheel must have some bearing on the direction relative to the amount of weight on it and that it must be the front wheel changing its angle of turn into the corner (maybe combined with a slight increase in wheel base) due to the front lifting more under acceleration. But... are we really talking about forward acceleration? At a higher constant velocity, won't the cornering forces increase and cause the suspension to compress more??? Argh!! Then again, maybe it is a matter of the rear wheel tracking a wider arc by itself, but, I don't know why. I'm going to have to sleep on it.

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If the suspension/geometry compresses/changes with more speed.... can we even use such a simple model at all to accurately predict lean angles?

Descriptive vs prescriptive?

We could figure out specific forces from lean angles and velocities at given radii, but... I'm not so sure about this entire model after all if the radius is an active/dynamic variable affected by mass... eh?

Oh my head is starting to hurt...

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Hey guys, very interesting post.

Isnt a wider arc caused by rolling friction (or the lack of) when the lean angle is maintained? When tires wear, small balls of rubber are tearing of the tire carcass and being shed. The more speed that is added, the more the tires slide. IE a car for example (no lean angle here) making a left turn, at full wheel lock, at a slow speed, will create an arc, 'X" amount in diameter. When speed is added the arc becomes larger relative to the speed. Rolling friction? Forward velocity?

I'm posing a question, not making an argument. I will not even try to pretend my mathematical abilities are close to the hieroglyphics I've seen in the previous posts.

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I'm not really sure, but next Friday we can do some experiments with the lean bike and find out

at least if Mr. Code will kindly turn his back and look somewhere else

Just want to leave enough time to go check out Mr. Barber's museum and get some pictures.

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I really want this thread to die now....but.....

I said the radius would increase if once leaned over you start feeding in more throttle. Seems reasonable. But one of the things brought up is what roll does the front wheel play while the bike is leaned over. So if I get ham fisted with the throttle and spin up the rear wheel the rear wheel will drift out on to a wider arc. What about the front wheel? Based on watching so really trick riding the front wheel stays on its line but with the rear wheel going out - the bike points more to the inside of the turn. I believe this is referred to as squaring off a turn and occasionally a highside.

I really can't wait till next weekend. I'm going to drove you guys nuts. I really want to spend some quality time with the lean/slide bike so I can get this in the real world.

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another point - reworking the formula and solving for r comes out something like:

r = v*v/(G*tan(50))

so clearly according to the formula if you increase velocity (but nothing else) you increase radius. Now if somebody could figure out a way to make local adjustments to G we'd be in business.

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Now if somebody could figure out a way to make local adjustments to G we'd be in business.

ROFL

I was just saying that this morning in reference to my perpetual motion free energy machine....

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Hey guys, very interesting post.

Isnt a wider arc caused by rolling friction (or the lack of) when the lean angle is maintained? When tires wear, small balls of rubber are tearing of the tire carcass and being shed. The more speed that is added, the more the tires slide. IE a car for example (no lean angle here) making a left turn, at full wheel lock, at a slow speed, will create an arc, 'X" amount in diameter. When speed is added the arc becomes larger relative to the speed. Rolling friction? Forward velocity?

I'm posing a question, not making an argument. I will not even try to pretend my mathematical abilities are close to the hieroglyphics I've seen in the previous posts.

One would need to be biased toward more front sliding as rear biased sliding would tend to tighten the arc... no?

Ok... I slept on it.

Then I went to work and messed around with one of the smaller dirt bikes in the showroom. I am not certain yet, but, it seemed that as I compressed the suspension from a leaned over position that the front wheel turned outward a bit from its starting position. I can't verbalize why yet (probably to do with the rake and where the tire contacts the ground, etc), but, I will spend more time with it tomorrow.

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OK... intuitively, I agree with 2bigalow. Thinking about understeer/oversteer. The front pushing into the turn. But once the bike is leaned over or the car is in the turn, without sliding, I think the car's turn radius will follow the angle of the front wheels and won't increase radius unless you slide some. So, I'm still leaning toward the front wheel angle changing somehow on the bike. We know that having the front compressed prior to turning creates "quicker" steering geometry, ie a faster flick due to a steeper steering head angle which gives us a more effective lever due to the contact patch being closer to the COG and less trail making the bike less stable or twitchy.

If we take the example of a chopper with really long forks and think about what happens when we turn the handlebar, it seems to me that the caster angle, that is the axis about which the wheel turns (nearly vertical in a car and less so on a sportbike) is even less vertical (closer to horizontal) on the chopper, eh? That is the wheel doesn't really turn exactly left or right, it also turns "up" and "down" somewhat. The front wheel is sort of 'heeled back' and, when turned, the contact patch actually moves off center with the bike and sort of rotates rearward on the wheel/tire wrt vertical. (I need to make a diagram here. The front wheel actually tracks a different arc than the rear. There are actually two wheel tracks running side by side through the corner.) So... here's the rub... when cornering forces are applied "vertically" through the bike, the "downward force" is not centered over the front wheel or more importantly the contact patch. It is actually on the inside of the contact patch. And it is applied at a forward angle as well due to the rake and more importantly applied forward of the contact patch due to the trail! Remember, trail is the steering head angle extending forward of a vertical line dropped from the axle, ie the contact patch when the bike is straight up and down.

What happens to the trail when you are leaned over? The front is compressed from cornering forces... less rake... less trail? Now what happens when you apply that compression force while leaned over? The angle of the force application is ahead of the contact patch. Hence, downward compression forces the wheel to attempt to straighten itself by applying force ahead of the contact patch. Or something like that. I think. Maybe. Or maybe I am sleep deprived and having another brain fart. I am too tired to edit this or read for mistakes. There may be glaring flaws of logic or dyslexic sentence construction. Does this make sense to anyone? Would you admit it if it did? lol

I'm going to bed now. I hope this doesn't read like the raving rantings of a madman idiot when I wake up tomorrow.

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And, taking it another step...

It has been suggested that when a rider rolls off the gas in a corner, the bike tries to countersteer itself up to less lean angle and that this is why a bike runs wide when we roll off in a corner.

However, given the line of thought in my last post, perhaps it is the forward weight transfer causing increased compression force at the front turning the wheel out to a wider angle. Hmmm...

*yawn*

*slurps coffee*

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OK, re-reading Chapter 13 of TOTW II...

"...once the bike is fully leaned into the corner, the rear end "steers" the machine." (p.58)

OK, so, adding more velocity, the bike wants to stand up due to increased cornering forces and we must apply force at the bars (or shift COG by hanging off) to maintain lean angle. But we have to lean even further than before to maintain radius. There seems to be a discrepancy. Something seems to be missing.

What happens when the rear wheel rolls faster? It must track a wider radius. Is it really "sliding"?

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Wouldnt the bars turn inward slightly since its the inside, leading edge, of the tire bitting the surface (pulling inward). IE a bicycle ghosting down the road. the bars will turn slightly to the inside by themselves. That tells me that leading edge and trail as you said, has something going on which is pushing the bike wide. Obviously this is without rear wheel power being put on the ground. Does this idea go out the window on the gas? IE a bike leaned over on the gas with the front wheel off the ground. Thus rolling on thru the whole corner.

I like to think of scenarios at extremes. So, What about negative trail? I mean ALOT lets say 50-70 mm behind the steering head. How does that effect the self correcting properties of a two wheeled vehicle? Again I'm posing a question. I dont have the mathematical abilities to put this on paper. I'm more of a build it and see what happens kind of guy. I can relate with your playing with a bike in the show room to see what happens

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what does this have to do with the 'real world'? Confidence. At least intellectually. If you KNOW that the bike can be leaned over to 50degrees and not fall over on you then you know how far you can lean the bike. knowing that then you can quick turn the bike that far and know that the bike won't fall down.

Tweek, I can understand your desire to comprehend the physics behind cornering, but I would be leery of gained confidence through that understanding .

If you KNOW that the bike can be leaned over to 50degrees and not fall over on you then you know how far you can lean the bike.

Refer to pages 18 & 19 in SS "Rider Physics"

I had a rider come to my Superbike School at Leguna Seca, and he crashed toward the end of the day. When we questioned him about what had happened, he told us, "I can't figure it out. I'm an engineering student, and my mother is an engineer, we're both heavy into physics. We sad down and figured this whole thing out by the physics of what happens while riding, and there was no way I could crash". Obviously, there was a flaw in his calculations. They didn't account for [bold]sampling[/bold]. This is another example of how [bold]a though or idea can determine what a rider does on the track.[/bold]

Just because "physics" say something can be done that doesn't necessarily mean that everyone can do it.

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bigdawg - agreed. However, if the physics say it can't be done - I'll let Stewy try it out first. If the math says its possible then I'm willing to work my way up to it.

The sucksville about doing the math is that, at least here, we really aren't making a accurate model of what is going on in the real world. We're not factoring in the suspension, the fact that the surface isn't perfectly smooth, the atmosphere (wind), tires, the rider's body position, etc, etc. All we have is a very simply model that says such and such a lean angle going around such and such a radius of turn will need about this much speed in order to balance the forces. I'm doubting my buddies at MSR are going to let me take my measuring tape out to measure the radius of all the turns on the 1.3 miles track. Oh, and that track is on the side of a hill so what about the camber of the turn? What about the fact that one of the narliest turns is crowned right at the apex? Way too complex to model, at least for me. based on Racer's posts though I'll expect a thesis any minute now.

At the end of the day I'm just killing time until the next track day or school. This helps my understanding which helps me a little bit when I jump on the bike, but at the end of the day - this is just for laughs.

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Wouldnt the bars turn inward slightly since its the inside, leading edge, of the tire bitting the surface (pulling inward). IE a bicycle ghosting down the road. the bars will turn slightly to the inside by themselves. That tells me that leading edge and trail as you said, has something going on which is pushing the bike wide. Obviously this is without rear wheel power being put on the ground. Does this idea go out the window on the gas? IE a bike leaned over on the gas with the front wheel off the ground. Thus rolling on thru the whole corner.

I like to think of scenarios at extremes. So, What about negative trail? I mean ALOT lets say 50-70 mm behind the steering head. How does that effect the self correcting properties of a two wheeled vehicle? Again I'm posing a question. I dont have the mathematical abilities to put this on paper. I'm more of a build it and see what happens kind of guy. I can relate with your playing with a bike in the show room to see what happens

OK, well I played with the KLX 110 instead of the CRF 80 today because it looked like it had more rake and might produce more pronounced results, which it seemed to. That said, we have indoor outdoor carpet and there are knobby tires on the bikes. Anyway, the more I played with the bike and looked at what happened and thought about what I wrote last night, the more sense it made.

I think if you apply force downward through the fork with "negative" trail, you will be forcing the wheel to turn more into the turn. I would define negative trail as a grocery cart with a front wheel turned backwards.

PS I'm not a big fan of using math to teach a concept. I think it acts like a wall to people who aren't skilled with it. My longstanding mantra is that I can teach the concept of any physics or engineering to anyone without the math easier than with it. That said, Tweek asked for a formula, so... I obliged. (That is the first time I have worked with math or equations of that sort since college... around 1986-87. That would be after slide rules and before cell phones and CD's. We had LCD digital watches and hand held scientific calulators, but no pc's or www.

Wait, not true. I used it a good bit when racing... like when deciding how much to drop my triple tree on my FZR's or add ride height on my GP bikes.

Oops, and I used it to win a bar room bet a couple of years ago ... I proved the altitude of geo-synchronous satellites was what I said it was ...very similar except using Pi and time instead of force, trig and vectors.

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