less than 15 degrees
somewhere around 20 degrees
somewhere around 30 degrees
somewhere around 40 degrees
more than 40 degrees
Shouldn't all you people be working or something?
Thats what I was saying, for 5khr, slow, tight turn, small radius to create 0.5G.
Can't say i've got my book on me, I remember we used engineering mechanics: dynamics, (meriam and kraige if i rememeber correctly)!! Vectors are king, who was that lecturer agian, simon B....someone i think, he had this slide he would put up "Only a fool doesn't draw a vector diagram!" haha
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Beer (appropriate name) and Johnson it was. You sound a little younger than me. We had Bryan Pidwerbesky who was the pavement guy take us for dynamics. He left about my 2nd Pro year.
I'm leaving this thread now (for about 2 seconds) cause it is getting a bit geeky (not that I can talk)
Cheers R
"The ultimate result of shielding men from the effects of folly is to fill the world with fools." - Herbert Spencer, English Philosopher (1820-1903)
Hmmm, I must say I'm not too happy about how you put it. There's a problem with causality!
If you're riding at speed "v" and want to make a corner with radius "R" - then you can calculate the necessary centripetal acceleration, "a", which is given by:
a=v^2/R (e.g. 0.5 G)
That's the necessary acceleration to realise a certain line. However, you don't achieve this acceleration by setting v and R to the right numbers. You achieve R for a given v if you apply the cornering acceleration a... I know, it's pretty pedantic - but the physical limitations are not set by v or R - but by how well your tyres grip (i.e. a).
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Assuming a Mike Hailwood riding style, ie upright, no hanging off, & no knee out, the width of the rear tyre will affect lean angle required for .5G as well. If say the rear tyre is a 150/17 and the bike is at full lean, which isn't the case here, the actual contact patch will be about 75mm to the side of the bike centreline which in itself will cause the vector of the combined forces to try and stand the bike back up. This will require a slightly greater lean angle to counteract.
If the resultant vector acted through the point of contact of the tyre when the bike was upright your formulae would be correct but they aren't taking into account that the contact point is actually a curved tyre and the point at which it contacts the road is displaced progressively to the side as the bike is leaned. The resultant vector still has to act through the centre of mass and the contact point of the tyre. The vector angle will stay the same but the bike will be leaning further because of the offset contact point of the tyre. I think!
The question was about "bike" lean angle, not "vector" angle.
Correct if you are looking at it from the perspective of how fast can I take a corner without falling off. This whole thread has been a bit backwards though, and we are looking at it from the perspective of how do I achieve horizontal acceleration vector of 0.5.
We didn't get a bike specified but if you are going to get that pedantic then the whole question is a bit farcical really. This is why I have stuck with resultant vector angles rather than bike angles. They are two very different things and unless we have details of the bike itself, the type of riding we are talking about (remember dirt/supermoto riders do it different to road riders) and the bike to rider relationship.
Lets go back to the pole's level of sensitivity. I'm guessing the answer is probably "somewhere around 20 degrees" for a road rider but I don't really know.
Cheers R
"The ultimate result of shielding men from the effects of folly is to fill the world with fools." - Herbert Spencer, English Philosopher (1820-1903)
Whats interesting is the spread of the pollsters.
Wonder what the spread would have been if the question was posed as to the maxium G force achievable on the bikes they ride. So many variables such as clearance, width of bike and rider and grip of tyres.
Wonder how many would say they pull more than 1G. And how many would say over 1.5G.
Exactly, hence the joke.
Say you pull 0.5G in corner on a moto-gp bike, then do the same on the munch mammut.
They tyres of the mammut would find it incredibly hard lean angles aside. Throw constant speed in there and its like rubbing the tyres down with vaseline.
I have never done a constant speed corner. I pray to the gods i never have too. May work on the new 2WD bikes though. depending on momentum balances.
Reactor Online. Sensors Online. Weapons Online. All Systems Nominal.
"AVGAS" lost me there. Surely if .5G is one vectors worth of force and gravity is the other vectors worth of force the resultant combined vector will be the same for both bikes, and not that terribly difficult to achieve either.
Humour me, I'm not an engineer.
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