Lean angle question.

[GSVR]

Heres a simple question. Motorcycle with heavy rims and tyres (High amounts of gyroscopic coupling going on) goes around a bend at 160 kph and leans 45 degrees. Now these wheels are taken off and some extremely light carbon rims are fitted that produce considerably less gyroscopic effect. The bike is ridden around the same bend at 160kph (all conditions the same, same gear, engine revs, COG when bike is vertical etc.) Only diffence is the wheels having considerably less gyroscopic coupling.

Its not a trick question only difference is the wheels. Everything else is the same.

Does the bike lean less, more, or the same amount? Feel free to give possible reasons for your answer.

I don't know the answer haven't tried to find out yet thats why I am asking.

[sAsLEX]

Leans the same. It is easier to lean and change direction though.


As the gyroscopic effect doesn't actually steer the bike the varying radii and direction of the tires does.

[k14]

It will also accelerate out of the corner faster due to less inertia in the wheels.

Oh and yes, I agree with saslex.

[Jantar]

There is no difference in the lean angle, just better all round handling.

[White trash]

Garry, just ask Glen if they were worth the money. We know that's what you're getting at

[yod]

I vote Garry for site Troller

[White trash]

Leans the same. It is easier to lean and change direction though.


As the gyroscopic effect doesn't actually steer the bike the varying radii and direction of the tires does.

No, gyroscopic effect doesn't alter the lean angle.

But put it this way. You've lightened the overall weight of the bike. But raised the center of gravity.

Nice question Garry, I like it.

[GSVR]

Garry, just ask Glen if they were worth the money. We know that's what you're getting at

Actually its not. Yes they great things to have but this is more to do with a Mechanical Technology book I'm looking at. It has a worked example of a motorcycle doing a 140kph 120 radius turn that is way over my head. But theres one bit that has me thinking. It says that the gyroscopic reaction tends to overturn the motorcycle outwards.

[GSVR]

But put it this way. You've lightened the overall weight of the bike. But raised the center of gravity.

I really don't want the COG changed so lets say its still the same (been fudged in some way)

[GSVR]

Well heres the question from the book it was printed in 1977:

A motor cyclist travels at 140 km/h round a cornerof 120 m radius. The cycle and rider have a mass of 150KG and their centre of gravity is 0.7 m above the ground level when machine is vertical. Each wheel is 0.6 m in diameter and the moment of inertia about its axis of rotation is 1.5 kg m2. The engine has rotating parts whose moment of intertia about their axis of rotation is 0.25 kg m2 and it rotates at five times the wheel speed in the same direction.

Find (a) the angle of banking so the there will be no tendency to sideslip, (b) the angle of the cycle and rider to the vertical.

The answer for (a) is 52 degrees and 5 minutes and (b) is 55 degrees and 33 minutes

To work out (a) they work out the forces acting on the bike - take the bikes weight in newtons ie 150kg x 9.81 = 1472 N. Then the centrifugal force - Mass of bike 150kg and the centrifugal force it creates going round a 120m radius at 140kph = 1890 N.
The resultant force from these two forces has to be perpendicular to the track surface.
(Tangent of 52 degrees and 5 minutes is 1.284) Vector diagram in book.

Goes on to work on the total gyroscopic coupling of the wheels and motor then uses the centre of gravity and forces acting about the point of contact (wheel and track) to arrive at 55 degrees and 33minutes.

So according to this the bike has to lean another 3 degrees and 28 minutes to overcome the gyroscopic forces wanting to "overturn the the cycle outwards" to quote the book.

All that really interests me at this point is the fact the bike has to lean more to overcome the gyroscopic forces. according to this book.

Mechanical Technology - D H Bacon and R C Stephens. Forth example in Section 16 Gyroscopic motion page 183.

I would have photoed the pages and posted but my camera is sitting in a ditch somewhere in Hawkes Bay.

[avrflr]

This effect is hard to work out intuitively, so it helps if you have a gyroscope to play with and see the effect. So one of the consequences of buying carbon wheels is that you won't have to lean the bike over so much for a given corner and speed. I hadn't thought about it before, that is interesting.

[GSVR]

This effect is hard to work out intuitively, so it helps if you have a gyroscope to play with and see the effect. So one of the consequences of buying carbon wheels is that you won't have to lean the bike over so much for a given corner and speed. I hadn't thought about it before, that is interesting.

Thinking of building a a self powered gyro. I have a lathe and heaps of high speed bearings as well as brushless outrunner motors and speed controllers that will get up to excessive rpm. Throw some high draw lithium cells into the mix and you have a radio controlled gyro lol. Na can't be bothered.

[avrflr]

Thinking of building a a self powered gyro. I have a lathe and heaps of high speed bearings as well as brushless outrunner motors and speed controllers that will get up to excessive rpm. Throw some high draw lithium cells into the mix and you have a radio controlled gyro lol. Na can't be bothered.

That would be the complicated way to go. I was thinking more along the lines of a spinning top or a frisbee on the finger.

The point is, when you push down on the outside of the frisbee (let's say) it will tilt in a direction 90 degrees from where your finger is. Now if you imagine the bike going around the corner then draw an axis perpendicular to the ground through the bike, you can imagine that you are twisting the bike around that axis as you turn the corner. Then you can picture the effect on your gyroscopes/wheels as you try to twist them around that vertical axis - they will resist by trying to stand up. As for how you work out the answer of 3 deg 28 minutes, that's just details.

[Mikkel]

Nice problem you lined up there...

If the bike is changing direction then you're changing the direction of the rotational momentum as well. To do that you need to apply an amount of torque orthogonal to both the rotating momentum and the desired change of momentum... But fuck if I can figure out how it all works out on top of my head.

[GSVR]

That would be the complicated way to go. I was thinking more along the lines of a spinning top or a frisbee on the finger.

The point is, when you push down on the outside of the frisbee (let's say) it will tilt in a direction 90 degrees from where your finger is. Now if you imagine the bike going around the corner then draw an axis perpendicular to the ground through the bike, you can imagine that you are twisting the bike around that axis as you turn the corner. Then you can picture the effect on your gyroscopes/wheels as you try to twist them around that vertical axis - they will resist by trying to stand up. As for how you work out the answer of 3 deg 28 minutes, that's just details.

I was thinking of building something that you could set the revs on accuarately while circling a set radius to achieve any lean angle you wanted by altering the gyro speed or the centrifugal force of the circling speed (radius and rpm). Would be an interesting display model.