Lean angle vs G-force

[Freakshow]

If 0.5G is going up does that mean they high sided it??

[Biggles2000]

To me the question was stated in the first post. 0 g riding straight and 0.5 addittional centrifugal loading g.

Cos(n)=1/1.5
n=48 deg lean angle

[ManDownUnder]

Surely it depends how fast you're going and the radius of the circle

at 5 kph you can't do it, and at 300kph it'll be on a considerably slighter angle (i.e. not so far off the vertical)

[Biggles2000]

You only have to know that the rider feels a 0.5g force in addition to gravity and that the bike ballances between being thrown out by centrifugal force and by falling in to the corner by the lean angle.
If you know the speed of the bike and the g force or centrifugal loading you can calculate the radius of the bend.

[Jantar]

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.

OK. So its not "Pulling" 0.5 G, its 0.5 G angular, so a lean angle (CoG to contact point) of 26.5 degrees.

[Biggles2000]

The g force is what the rider feels so in this case he is experiencing 1.5g. The hypotenuse is 1.5g and the adjacent is 1g.

[The Stranger:Me]

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

Surely you aren't referring to correcting your overly complex and total bullshit description of countersteering are you?

[GSVR]

Surely you aren't referring to correcting your overly complex and total bullshit description of countersteering are you?

It works for me until someone educates me otherwise.

[*caution*]

Surely it depends how fast you're going and the radius of the circle

at 5 kph you can't do it, and at 300kph it'll be on a considerably slighter angle (i.e. not so far off the vertical)

you don't need those to work out the lean angle, if your leaning at 26.6 degrees at 5kph, while technically possible, it would quite a tight turn, hmm, give me a minute to work out the bend raidus required, at 300kph i don't know my bike can't go that fast!!!!!

hang on I think I might have screwed this up royally along with everyone else who worked out 26.6 degrees....

will report back.... off to study physics......

EDIT: no its alright, just overthinking....

[cooneyr]

The g force is what the rider feels so in this case he is experiencing 1.5g. The hypotenuse is 1.5g and the adjacent is 1g.

Ah Na. GSVR clarified - 0.5G horizontally so....

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.

As Jantar said the resultant vector is (1^2+0.5^2)^0.5 = 1.118G leading to

But the resultant of that vector is 1.118 G.

the resultant vector angle is therefore being (allowing for rounding of the 1.118 which is actually 1.11803398874989 to as many DP as excel will give me)

COS(n) = 1/1.118 hence n = 26.56 or

TAN(n) = 0.5/1 = hence n = 26.57 or

SIN(n) = 0.5/1.118 hence n = 26.57

Therefore if the horizontal vector is 0.5G we can definitively say the resultant vector angle is 26.57 degrees from vertical. How you achieve the resultant vector of 1.11803398874989 at this angle is entirely up to you - inline with the bike or hanging off the side. By hanging of the side you can achieve the resultant vector with less lean angle on the bike however in either situation because the horizontal vector is fixed at 0.5G the speed will be the same whether hanging of or not. The only thing that hanging off changes is the lean angle of the bike - the system lean angle stays the same and hence the speed is the same. This then leans to the well understood fact that you try to go faster (increase magnitude and angle of the resultant vector) and bits of the bike touchdown you need to get your arse of the side and the bike more upright.

Cheers R

[Badjelly]

Yep, the correct answer, to a first approximation, is arctan(0.5) = 26.6 degrees. And we can't give a better answer because there's not enough info. It's a pity the poll doesn't offer "about 25 degrees" as an answer, but failing that I'll go for 30.

[DingoZ]

The answer is ...... USD FORKS gifts all bikes with greater lean angle and cornering speed...

done.......where's the prize...








Pisstake Folks - no need to incerate a poster

[avgas]

Hold up where do you mean 0.5G? Rotational G? Lateral G? Vertical G?
Or do you mean G at the tyre (i.e. the grip holding the balance so that you dont fly away like a squid)?
How long is a piece of string?



*Edit*
hahah classic, just got the joke - going constant around a corner. Classic

[cooneyr]

you don't need those to work out the lean angle, if your leaning at 26.6 degrees at 5kph, while technically possible, it would quite a tight turn, hmm, give me a minute to work out the bend raidus required, at 300kph i don't know my bike can't go that fast!!!!!

hang on I think I might have screwed this up royally along with everyone else who worked out 26.6 degrees....

will report back.... off to study physics......

Its not about the speed but rather the vectors. If you want to achieve a 0.5G horizontal vector you can either travel fast around a large radius or relatively slower around a small radius. Time to crack out the radial vector math me thinks - ummm where the hell is that. Damn - just found my first year Vector Mechanics for Engineers (Dymanics) text. Will have to read an remember how the hell it all goes later - was about 12 years ago I last looked at that text

Cheers R

[Badjelly]

Hold up where do you mean 0.5G? Rotational G? Lateral G? Vertical G?

Rotational G???

The bike is going around a corner of constant radius at constant speed such that the centripetal (or lateral) acceleration of centre of mass of the bike + rider is 0.5 G. This requires a lateral force at the tyre contact patches totalling (rider+bike mass) x 0.5 G.

It's a level road (he didn't say otherwise) so weight on the road is (rider+bike mass) x 1.0 G.

We want a lean angle such that the force vector from the tyre goes through the centre of mass of the bike + rider, otherwise we're in trouble. This is arctan(0.5) and total force through the contact patch is mass times sqrt(1.0^2+0.5^2) = 1.18 G.

The biggest confusing factor is the torque required to change the angular momentum of the wheels, though I *suspect* this just affects the allocation of load to the front and rear wheels and doesn't affect the answer.