Quote:
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.
|
I didn't expect to be given homework when I signed up to UKMOC, Rich!
Just to put calculations to Darkness's detailed reply; you can still calculate the various forces involved at different lean angles using trigonometry.
Side 'a' represents the weight of bike and rider
Side 'b' represents the lateral cornering force
Side 'c' represents the resultant force acting on the suspension.
The angle
theta represents 90° minus the lean angle.
Using trig:
sin
theta = a/c
cos
theta = b/c
We know three elements of the triangle:
• 'a' = 1
• The angle between a and b is 90°
• The angle
theta is 24° (i.e. 90° - 66°)
So, as sin 24° is 0.4067 then 'c' = 1/sin
theta =1/0.4067 =
2.4588
And as cos 24° is 0.9135 then 'b' = c x cos
theta = 2.4588 x 0.9135 =
2.2462
So the lateral cornering force = 2.2462g and the effective load on the suspension is 2.4588g.
The simple way to find the cornering force is just to look up the tan of the lean angle i.e. the tan of 66° is 2.246 and this is effectively what this graph shows (again, courtesy of Tony Foale)
I'm just off now for a couple of Anadin
