Lean angle vs G-force

**xwhatsit** · 8th February 2008, 12:12

I'm no great physics student, but your lean angle would get greater the longer you cornered, right? Until you fell over? As, on a bike, there's no wheels on the other side to hold you upright.

Unless you're not talking about remaining upright as you say, and instead are talking about leaning into the corner?

**Jantar** · 8th February 2008, 12:15

Originally Posted by Mikkel

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. ....

Correct, but if you are cornering at 0.5 G angular acceleration, then you are pulling 1.118 G. That is the combined vector of 1G vertical plus 0.5 G horizontal.

Therefore Johan is asking the correct question. Does GSVR actually mean pulling 1.5G, or is it 0.5G angular?

If the first then its 56.3 degrees, if the second then its 26.5 degrees.

**speedpro** · 8th February 2008, 12:15

Wheel width is a big factor as well because as you lean with wide tyres the contact patch moves away from the centre line of the bike requiring more lean to balance the forces generated by cornering than a similar bike with narrower tyres and a contact patch relatively in line with the centre line of the bike, maybe.

**Usarka** · 8th February 2008, 12:18

He's talking about G-force, not gravitational acceleration. Isn't he?

Isn't it the same thing? 1 g force = 9.8 m/s² When you are cornering in a a F1 the g force is lateral, if 1g then the sideways force is equal to the downwards force gravity is exerting.

I reckon the profile of the tyres is a big factor. And the setup of your forks (especially USD)

Fucked if i know, i'm drunk.

**hazard02** · 8th February 2008, 12:20

I'm going with 26.57 degrees (me thinks).

(!)Bonus vector diagram:

**Jantar** · 8th February 2008, 12:25

Originally Posted by hazard02

...(!)Bonus vector diagram:

But the resultant of that vector is 1.118 G.

***caution*** · 8th February 2008, 12:26

Originally Posted by skelstar

He's talking about G-force, not gravitational acceleration. Isn't he?

the acceleration due to gravity is 9.81.... ms^2 as someone mentioned, otherwise known as 1G, people talk about "G" forces because you immediatly have something to compare it with and have feel for what the acceleration might be, so if people say a jet fighter might turn at 78.48 ms^2, you'd be like huh? if they said 8G you'd be like whoa. maybe.....

Now wheres that thread about the G-Spot....

***caution*** · 8th February 2008, 12:30

Originally Posted by hazard02

I'm going with 26.57 degrees (me thinks).

(!)Bonus vector diagram:

HI Five man, thats what I got.

Originally Posted by hazard02

But the resultant of that vector is 1.118 G.

It would have to be if you wanted to turn, note that the resultant is not acting straight down, but inward, thats the cornering force plus the weight force.

**hazard02** · 8th February 2008, 12:32

Originally Posted by Jantar

But the resultant of that vector is 1.118 G.

Sure is. Adding some lateral g-force doesn't negate the g-force of gravity. The rider experiences a resultant g-force of 1.118G at an angle of 27 degrees.

EDIT: beaten : (

RE-EDIT: Hi five! Great minds clearly think alike.

**Mikkel** · 8th February 2008, 12:38

Originally Posted by *caution*

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!!

Which is why MotoGP riders need to hang off the bike to increase the effective lean angle. Remember that the lean angle of the bike is not the same as the lean angle of the system - only thing that matters here is the line that goes from the contact patch through the centre of mass.

**Mikkel** · 8th February 2008, 12:43

Originally Posted by *caution*

It would have to be if you wanted to turn, note that the resultant is not acting straight down, but inward, thats the cornering force plus the weight force.

Sitting straight up and down you would perceive gravity to be pulling you straight into your seat at 1.118 G.
The perceived forces are opposite to the acting forces as a general rule.

**GSVR** · 8th February 2008, 12:50

Originally Posted by Jantar

Correct, but if you are cornering at 0.5 G angular acceleration, then you are pulling 1.118 G. That is the combined vector of 1G vertical plus 0.5 G horizontal.

Therefore Johan is asking the correct question. Does GSVR actually mean pulling 1.5G, or is it 0.5G angular?

If the first then its 56.3 degrees, if the second then its 26.5 degrees.

Knew it would get messy.

The G force is measured exactly perpendicular (90 degrees) to Gravity and from the centre of the arc (radius) of the turn.

***caution*** · 8th February 2008, 12:50

Originally Posted by Mikkel

Which is why MotoGP riders need to hang off the bike to increase the effective lean angle. Remember that the lean angle of the bike is not the same as the lean angle of the system - only thing that matters here is the line that goes from the contact patch through the centre of mass.

Exactly right mate, that angle would be the lean angle of the centre of mass of the rider plus bike, which wouldn't be the centre of the bike if he was leaning off, and as you get closer to the ground he can only lean off so far....

***caution*** · 8th February 2008, 12:53

Originally Posted by Mikkel

Sitting straight up and down you would perceive gravity to be pulling you straight into your seat at 1.118 G.
The perceived forces are opposite to the acting forces as a general rule.

yeah its always a bit confusing in physics if your talking about the acting or resultant forces, got many a question wrong by mixing up my positives and negatives!!! The force you feel is the seat pushing back at you as you try to accelerate down on it...

**cooneyr** · 8th February 2008, 12:55

You dont want to increase the resultant vector angle because once the horizontal vector component starts to get close to 1g bad shit starts to happen. i.e. things start to let go. Simple analogy - those "fancy" brake efficiency testers at the likes of VTNZ. They are just simply the percentage of the vertical force of the car (i.e. weight times gravity) that is converted into horizontal force. 100% is almost impossible. F1 cars are kinda special due to fancy tyres, wings and fancy suspension.

By the rider leaning off further then they can hold the bike more up right for the same effective vector angle. This means you can pull more horizontal G (go faster because radial acceleration is dependent on speed around the curve) for the same bike lean angle (not system lean angle). Bike angle less i.e. more upright with the rider hanging of the side and the system angle remains the same for any given speed or put another way - bits of the bike don't hit the ground as you go faster

Cheers R