Suspension and cornering forces

[Mr Gazza]

EDIT (slob): Moved/copied from 350TSS’s rebuild thread, so apologies if it seems to start a bit randomly /EDIT



Why have you fitted a fore and aft baffle?
I don't think fuel sloshes from side to side. Or does it? What forces move it sideways?
I sit on my bike and feel no sideways force. Any cornering force is countered by leaning.

I think there are large and more undesirable fore and aft forces when braking and accelerating.

[Mr Gazza]

Why have you fitted a fore and aft baffle?
I don't think fuel sloshes from side to side. Or does it? What forces move it sideways?
I sit on my bike and feel no sideways force. Any cornering force is countered by leaning.

I think there are large and more undesirable fore and aft forces when braking and accelerating.

[350TSS]

I am not sure I agree with you - if you consider taking a chicane for example, I think that the input force to make the bike change direction occurs before the bike is subject to the centrifugal (or is it centripetal force) when the bike is in that part of the chicane. I think it unlikely that the fuel level within the tank stays parallel with the bottom of the tank throughout the manoeuvre.
I may be wrong and I may have wasted a day of my life.
In any event 3 x cross ways baffles are planned for the tank top to stop the sloshing forward under braking

[350TSS]

I am not sure I agree with you - if you consider taking a chicane for example, I think that the input force to make the bike change direction occurs before the bike is subject to the centrifugal (or is it centripetal force) when the bike is in that part of the chicane. I think it unlikely that the fuel level within the tank stays parallel with the bottom of the tank throughout the manoeuvre.
I may be wrong and I may have wasted a day of my life.
In any event 3 x cross ways baffles are planned for the tank top to stop the sloshing forward under braking

[Darkness]

Originally Posted by 350TSS

I am not sure I agree with you - if you consider taking a chicane for example, I think that the input force to make the bike change direction occurs before the bike is subject to the centrifugal (or is it centripetal force) when the bike is in that part of the chicane. I think it unlikely that the fuel level within the tank stays parallel with the bottom of the tank throughout the manoeuvre.
I may be wrong and I may have wasted a day of my life.
In any event 3 x cross ways baffles are planned for the tank top to stop the sloshing forward under braking

I believe you are right. The steering input causes the bike to tilt and rotate about the centre of gravity. Depending where the fuel is relative to the centre of gravity the fuel could just be subjected to the tank rotating, but is more likely to think the tank has just lurched to left (Or right) as well.

It’s probably more noticeable on a very tall trailee, or enduro, but the more the fuel is able to rush to the sides of the tank is likely to add a vagueness to the steering.

[350TSS]

I think the bike rotates around the tyre contact patch and the bottom of the tank is about 800mm above that on a M900. A lot more on a Cagiva Elefant for example. The greater the distance betweem the contact patch and the bottom of the tank the more fuel moves across the tank. I do not think that a bike pulls 1 G when cornering so the fuel level will tend to stay level with the ground not level with the bottom of the tank.
I could be talking absolute bollocks!!!!

[Darkness]

Originally Posted by 350TSS

I think the bike rotates around the tyre contact patch and the bottom of the tank is about 800mm above that on a M900. A lot more on a Cagiva Elefant for example. The greater the distance betweem the contact patch and the bottom of the tank the more fuel moves across the tank. I do not think that a bike pulls 1 G when cornering so the fuel level will tend to stay level with the ground not level with the bottom of the tank.
I could be talking absolute bollocks!!!!

Not quite.

The bike and rider continue in a constant state until acted on by a force.

To go round a right hand bend the rider initially steers left to make the bike lean. The steering is then pretty neutral till you want the bike to stand up and go straight again. That is done by oversteering tonthe right so that the wheels come back underneath the bike.

The force acting on the bike is at ground level, but the bikes inertia is acting at the centre of gravity. The difference between these gives a lever arm and generates rotation. If the bike rotated about the contact patch there’d be nothing to make the bike lean.

The fuel isn’t a lot higher than the centre of gravity of the bike and ride, but there would be sideways motion as well as rotation.

So, in summary, yes there is an effect, though possibly not as great as you first thought.

[Luddite]

One for the mathematicians/physicists amongst you! (45° = 1g)



Courtesy of Tony Foale again "Motorcycle handling and chassis design".

[Luddite]

Originally Posted by 350TSS

Re the above I do not understand why R =1.414 - what is the origin of the equation? why is it the square root of 2 x weight?

I did say this was one for the mathematicians, (one of which I am not!).

To determine the resultant magnitude of two (and only two) vectors that make a right angle to each other, the Pythagorean theorem can be used.

So in the above example, the hypotenuse (R) is found by finding the square root of the sum of the squares of the other two sides, (i.e. the vertical (v) and horizontal (h) components).

So:
R = Square root of v² + h²

R = Square root of 1² + 1²

R = Square root of 2

R = 1.4142

Who knew bikes could be so boring?! (But you did ask!)

[utopia]

Going back to the cornering discussion ....
Surely the bike must rotate laterally about the tyre contact patch, not the centre of gravity ....?
In order for it to rotate about the C of G, the contact patch would have to move laterally on the road, would it not ...??

[Capo]

Originally Posted by utopia

Going back to the cornering discussion ....
Surely the bike must rotate laterally about the tyre contact patch, not the centre of gravity ....?
In order for it to rotate about the C of G, the contact patch would have to move laterally on the road, would it not ...??

^^^^ Correct

[350TSS]

Re the above I am still troubled by the TF diagram. I am not an engineer nor am I a mathematician (O level just 51 years ago).
What caused me to be troubled was the news that Marc Marquez has been recorded as taking a corner with a record 66 degrees of lean and I wondered what the cornering force could be, so I thought I would try to calculate it. Would the force be higher or lower once 45 degrees of lean was passed?
As I thought about it further I was also worried by the statement that the diagram only worked for 45 degrees. One should be able to calculate at 5 degree increments and plot a graph.
To help me consider the issue I drew a quarter of a circle and plotted 10 degree increments including one last position at 66 degrees. It is a frighteningly long way over.
I also thought about the effect the rider would have by shifting weight off and inboard of the machine as the corner is taken. I think a MotoGP bike has a mass of about 135 kg and Marc fully kitted probably weighs about 65 kgs so all up about 200 kgs.
Looking at the TF diagram in the vertical position the force is through the rider to the tyre contact patch and is the Wt in kgs. I could not work out the position anywhere between the tyre contact patch and the top of the rider’s head which is the correct position to measure from. I concluded that it could only theoretically be measured from the centre of gravity of the rider/ bike combination and for the purpose of simplification of the calculation one would have to assume that the rider remained in the same place on the seat.
Also for the sake of simplicity and my other severely overloaded brain cell I have assumed that the centre of gravity of the Marc and bike combination is 750mm above the tyre contact patch in the vertical position. As the bike leans the CofG of the bike /rider combination prescribes an arc 750mm distant from the tyre contact patch with increasing lean angle reducing the height of the C of G above the ground level.
Mr Foale’s diagram has the radius of this arc as the hypotenuse of a right angle triangle and in his case the answer is always the square root of 2 x the weight in kgs. Force is not measured in kgs but rather in Newtons.
I am obviously missing something
Discuss?

[Darkness]

Originally Posted by utopia

Surely the bike must rotate laterally about the tyre contact patch, not the centre of gravity ....?

No. Newton’s first law is ‘an object either remains at rest or continues to move at a constant velocity, unless acted upon by a force’, so ignoring friction and air resistance a moving bike will just continue straight down the road till you (Or something else) does something. If you want to turn a corner you have to do something with the steering. For a right hand bend you need to get the bike leaning, so you have to steer LEFT. Only long enough to get the bike leaning. Effectively that steers the wheels out from under you. If you want to check this, try steering a heavy bike with very short clip ons. The steering effort on the bars becomes quite noticeable.
Once the bike is leant over you don’t have to steer. You could go “no hands” if you wanted. The bike will just carry on in a circular curve until you change things again.
To pick the bike back up to vertical you have to steer the front wheel back under you. The back wheel just follows it.

If you want a practical demonstration, try making tyre tracks with a push bike when there are some puddles. You can see that the right corners start with the front wheel steering to the left, then swinging round to the right once you are leaning.

Originally Posted by utopia

In order for it to rotate about the C of G, the contact patch would have to move laterally on the road, would it not ...??

Yes. It does.

[slob]

Originally Posted by Darkness

...Once the bike is leant over you don’t have to steer. You could go “no hands” if you wanted. The bike will just carry on in a circular curve until you change things again...

I’ve watched Simon Crafar demonstrate this through the penultimate corner at Aragon, approaching three digit speeds on a GSXR1000 with his hands above his head... mental

[Darkness]

Originally Posted by 350TSS

Re the above I am still troubled by the TF diagram. I am not an engineer nor am I a mathematician (O level just 51 years ago).
What caused me to be troubled was the news that Marc Marquez has been recorded as taking a corner with a record 66 degrees of lean and I wondered what the cornering force could be, so I thought I would try to calculate it. Would the force be higher or lower once 45 degrees of lean was passed?
As I thought about it further I was also worried by the statement that the diagram only worked for 45 degrees. One should be able to calculate at 5 degree increments and plot a graph.
To help me consider the issue I drew a quarter of a circle and plotted 10 degree increments including one last position at 66 degrees. It is a frighteningly long way over.
I also thought about the effect the rider would have by shifting weight off and inboard of the machine as the corner is taken. I think a MotoGP bike has a mass of about 135 kg and Marc fully kitted probably weighs about 65 kgs so all up about 200 kgs.
Looking at the TF diagram in the vertical position the force is through the rider to the tyre contact patch and is the Wt in kgs. I could not work out the position anywhere between the tyre contact patch and the top of the rider’s head which is the correct position to measure from. I concluded that it could only theoretically be measured from the centre of gravity of the rider/ bike combination and for the purpose of simplification of the calculation one would have to assume that the rider remained in the same place on the seat.
Also for the sake of simplicity and my other severely overloaded brain cell I have assumed that the centre of gravity of the Marc and bike combination is 750mm above the tyre contact patch in the vertical position. As the bike leans the CofG of the bike /rider combination prescribes an arc 750mm distant from the tyre contact patch with increasing lean angle reducing the height of the C of G above the ground level.
Mr Foale’s diagram has the radius of this arc as the hypotenuse of a right angle triangle and in his case the answer is always the square root of 2 x the weight in kgs. Force is not measured in kgs but rather in Newtons.
I am obviously missing something
Discuss?

The neat thing about vectors is that they have length and direction.
The root two only applies to 45 degree triangles. In this case the angles are different.
For Mr Marquez, if you draw a triangle with horizontal base line, a vertical side and a diagonal line at 66 degrees from vertical.
The vertical line shall represent the weight of bike and Marc, say 200 Kg force. The horizontal line represents the turning force and the diagonal the force in the suspension.

For this 66 degree rightangled triangle, the hypotenuse is 2.459 times longer than the vertical short side, so Marc is pulling 2.459 G around that corner, plus any acceleration or braking.

The suspension thinks that he and the bike now weigh 491.7Kg.