Lean angle vs G-force

**GSVR** · 8th February 2008, 11:07

If you pull 0.5 G going around a constant corner sitting perfectly upright (inline with bike) how far would your bike lean ( 0 being vertical)?

I expect the usual lame jokes or added complexity from people that are into doing that.

***caution*** · 8th February 2008, 11:23

Its obviously the inverse tan of 0.5,

26.6 degrees?

**onearmedbandit** · 8th February 2008, 11:24

Where is the option for "I wouldn't have the foggiest"? I'm expecting a point to this, when do we find out?

(btw, I could probably google the question but then that wouldn't be my answer.)

**Forest** · 8th February 2008, 11:35

Originally Posted by GSVR

If you pull 0.5 G going around a constant corner sitting perfectly upright (inline with bike) how far would your bike lean ( 0 being vertical)?

I expect the usual lame jokes or added complexity from people that are into doing that.

There is no answer to this question, since the answer depends on a number of factors which you haven't given us.

**LilSel** · 8th February 2008, 11:37

I have totally no idea... as OAB said... 'wheres the option for not the foggiest' lol...

**GSVR** · 8th February 2008, 11:39

Originally Posted by Forest

There is no answer to this question, since the answer depends on a number of factors which you haven't given us.

What other factors? This should be interesting.

***caution*** · 8th February 2008, 11:43

Originally Posted by Forest

There is no answer to this question, since the answer depends on a number of factors which you haven't given us.

Everything your not given cancels out if you do the calculation, assuming i'm right and not making an egg of myself.

**johan** · 8th February 2008, 11:49

Originally Posted by GSVR

If you pull 0.5 G going around a constant corner sitting perfectly upright (inline with bike) how far would your bike lean ( 0 being vertical)?

0.5 G would be negative G. Did you mean 1.5 G?

**Usarka** · 8th February 2008, 11:52

Originally Posted by johan

0.5 G would be negative G. Did you mean 1.5 G?

Huh?

....

**skelstar** · 8th February 2008, 11:57

Originally Posted by johan

0.5 G would be negative G. Did you mean 1.5 G?

Negative G is floating up. 0.5G just means 'half weight' blah blah.

Yeah, 0.5G in which direction Garry? At rest the bike is exerting 1G down through the wheels. Could only manage less than that if you go over a crest or something, so the answer is most likely 'vertical'.

**johan** · 8th February 2008, 11:58

Originally Posted by skelstar

Negative G is floating up. 0.5G just means 'half weight' blah blah.

Ah ye, you're right, my bad.

**Mikkel** · 8th February 2008, 12:02

Jesus christ...

A G is an acceleration equal to roughly 9.8 m/s^2.

If you're cornering at 0.5 G it means that the rate at which you change your direction of travel is 4.9 m/s^2. If you don't move your weight around (cause no lateral displacement of your centre of mass) the angle in relation to vertical should be found by taking the inverse tangens function of 2. But right now my calculator is out of batteries.

**Devil** · 8th February 2008, 12:06

Wheelbase and centre of gravity are factors...

***caution*** · 8th February 2008, 12:09

Originally Posted by skelstar

Negative G is floating up. 0.5G just means 'half weight' blah blah.

Yeah, 0.5G in which direction Garry? At rest the bike is exerting 1G down through the wheels. Could only manage less than that if you go over a crest or something, so the answer is most likely 'vertical'.

I think hes referring to the cornering G (acceleration toward the centre of the arc), horizontal G, rather than the vertical G's

This got me calculating......a MotoGP bike would need to lean to about 78 degrees to keep up with the cornering G's of an F1 car (4-5G's of the top of my head??), you'd need some damn grippy tires and alot of clearance to do that!!