MotoGP - coefficient of friction 55¦=1.4 2

[crysta1c1ear]

During the TV commentary of the penultimate MotoGP, I heard them say that the bikes were leaning over at 55 degrees in one of the bends.

My ears pricked up, as I took that to be a statement of the coefficient of friction of the tyres: a figure I'd been interested in but didn't expect to hear.

Today I got out my windows calculator, typed in 55 and hit the TAN button, and got 1.4. My reasoning is that
the centripetal force is acting horizontally (opposite side) and gravity is acting vertically (adjacent side) in my triangle of forces, where the bike is leaning at 55 degrees, presumably to the vertical.

I had expected the number to be closer to 1.2. I know the 55 degrees figure will have been rounded. TAN(50) is 1.2 and TAN(52.5) is 1.3. I figure they could have rounded anything above 52 and a half degrees to 55 so maybe I should only trust the figure as far as saying the coefficient of friction is over 1.3.

Or maybe the bend was banked and I hadn't noticed.

Is 1.4 coefficient of friction for a motoGP bike's tyres and a 55 degree lean on a horizontal bend feasible?

 

[crysta1c1ear]

I just came across an illustrative picture, as I was looking for a head on picture where I might see angles for myself so that I could "measure" things.

 

[GregLocock]

Bear in mind that the thrust from the rear wheel contributes to the centripetal force if beta is not zero. I suppose you could argue that this still uses the friction ellipse (if that is the way you think), but I suspect a bike tire has more longitudinal grip than lateral.

Not that 1.4 is unbelievable - a street car tyre can produce 1.2 under braking.

If we're lucky Ben will join in. but he'll have to kill us afterwards.



Cheers

Greg Locock

SIGlease see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[crysta1c1ear]

Greg
I understand what you are saying. I am assuming that they lean the bike over further as the brakes are reduced and that the bike stands up again as power is brought in. With that assumption there will be little power fed to the wheels when cornering hardest.

I'm trying really to analyse a hypothetical instant at the end of braking and the start of acceleration, which is neither one nor the other.

I suspect a bike tire has more longitudinal grip than lateral.

I'd agree with that.

convert Biaggi.jpg -rotate 51 Traction51.gif

I ran a few command lines on my computer. To be honest, it is hard to tell where the centre of mass is with the riders leaning further than the bikes, and with the contact patch being at the side of the bike rather than underneath in the rotated pictures.

Once again, I have a picture on my PC which won't display 99% of the time, because I won't be logged on. Sorry. It's just the same picture rotated 51 degrees. Presumably today's tyres grip a little better.

 

[oab]

crysta1c1ear
"Is 1.4 coefficient of friction for a motoGP bike's tyres and a 55 degree lean on a horizontal bend feasible?"

Well when you see them doing it,, I guess it is.lol
I am probable not going to explain this too well, but.

When you bank a bike over, you move the center of gravity (mainly the rider and fuel tank) to the inside of the track of the wheels. The centrifugal force that is acting on this mass is vectored down through the frame to increase the down force on the tyres.

You can feel this happen in a corner, as the bike compresses the suspension down. So the harder you corner the better you grip.

Harvey.

 

[ubrben]

Hi Greg – you’re right I will have to kill you all

Before answering the question it’s important to know how this data’s being generated. Because of the lean angle it’s impossible to conventionally measure lateral G. Integrating roll rate to give lean angle to resolve accelerometer readings is pretty crap due to drift.

As a result everyone uses GPS to calculate a lateral acceleration in the plane of the track surface – however this is in error if there’s banking – same as any lateral accelerometer in a car – generally requires a track walk on setup day to make sure you know where the banking is.

The lean angle you see on the TV is simply arctan(lateral G) and that’s the same as we used on the data logging to analyse tyre performance.

The short answer to the OP’s two questions is yes and yes. Lateral G calculated by the method described above tends to reach a steady state average of around 1.5G in mid-corner (56 degrees lean angle) and transient spikes of up to 1.8G (61 degrees lean angle) are routinely seen in fast direction changes.

Those numbers are entirely consistent with lateral G traces recorded directly with accelerometers on cars with no downforce and sticky race tyres – e.g. Formula SAE and are basically indicative of the friction coefficient that is achievable with modern soft racing slick compounds.

Ben

 

[ubrben]

Re-read my post and had some immediate other thoughts.

The riders (Carlos Checa and Colin Edwards being the best I have data for) tend not to use the absolute limit laterally in a corner – they tend to hit foot pegs on the ground, etc before that. As a result the G-G diagram as measured is very square edged with up to +- 0.3G Longitudinal G at maximum lean angle carried by the rider.

The main benefit of Q tyres is at an increased friction limit they can carry the same lean angle and accelerate harder and trail brake deeper. The max lateral G rarely goes up by as much as the longitudinal G with a Q.

The other point Greg made was about longitudinal capability – unfortunately this is difficult to quantify using track data because the bike is wheelie limited on traction to about 1G. You do see heavy braking in a straight line at high Long G with the rear wheel in the air, but that’s with the front tyre. Traction and the bulk of the cornering coming from the rear tyre.

Without flat-trac data it would be difficult to be sure but in terms of contact patch length:width ratio I suspect Greg’s right.

Ben

 

[crysta1c1ear]

Thanks Ben.