Lean angle vs G-force

[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.

[*caution*]

Its obviously the inverse tan of 0.5,

26.6 degrees?

[onearmedbandit]

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]

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]

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

[GSVR]

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*]

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. :confused:

[johan]

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]

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

Huh?

....

[skelstar]

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]

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


Ah ye, you're right, my bad.

[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. 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]

Wheelbase and centre of gravity are factors...

[*caution*]

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!!

[skelstar]

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

[xwhatsit]

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]

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]

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]

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]

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

(!)Bonus vector diagram:

[Jantar]

...(!)Bonus vector diagram:
But the resultant of that vector is 1.118 G.

[*caution*]

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*]

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

(!)Bonus vector diagram:

HI Five man, thats what I got.


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]

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]

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]

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. :yes:

[GSVR]

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*]

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*]

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. :yes:

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]

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 :D

Cheers R

[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]

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 :D

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.

[Badjelly]

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.

Sorry, cooneyr already said all that.

[onearmedbandit]

Shouldn't all you people be working or something?

[cooneyr]

Shouldn't all you people be working or something?

Remembering the "old" days at Eng school is much more fun than work :D

Cheers R

[*caution*]

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 :D

Cheers R

Thats what I was saying, for 5khr, slow, tight turn, small radius to create 0.5G.

Can't say i've got my book on me, I remember we used engineering mechanics: dynamics, (meriam and kraige if i rememeber correctly)!! Vectors are king, who was that lecturer agian, simon B....someone i think, he had this slide he would put up "Only a fool doesn't draw a vector diagram!" haha

[Badjelly]

Shouldn't all you people be working or something?
I'm *at* work, will that do?

[cooneyr]

Thats what I was saying, for 5khr, slow, tight turn, small radius to create 0.5G.

Can't say i've got my book on me, I remember we used engineering mechanics: dynamics, (meriam and kraige if i rememeber correctly)!! Vectors are king, who was that lecturer agian, simon B....someone i think, he had this slide he would put up "Only a fool doesn't draw a vector diagram!" haha

Beer (appropriate name) and Johnson it was. You sound a little younger than me. We had Bryan Pidwerbesky who was the pavement guy take us for dynamics. He left about my 2nd Pro year.

I'm leaving this thread now (for about 2 seconds) cause it is getting a bit geeky (not that I can talk) :rofl:

Cheers R

[Mikkel]

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.

Hmmm, I must say I'm not too happy about how you put it. There's a problem with causality!

If you're riding at speed "v" and want to make a corner with radius "R" - then you can calculate the necessary centripetal acceleration, "a", which is given by:

a=v^2/R (e.g. 0.5 G)

That's the necessary acceleration to realise a certain line. However, you don't achieve this acceleration by setting v and R to the right numbers. You achieve R for a given v if you apply the cornering acceleration a... I know, it's pretty pedantic - but the physical limitations are not set by v or R - but by how well your tyres grip (i.e. a).

;)

[speedpro]

Assuming a Mike Hailwood riding style, ie upright, no hanging off, & no knee out, the width of the rear tyre will affect lean angle required for .5G as well. If say the rear tyre is a 150/17 and the bike is at full lean, which isn't the case here, the actual contact patch will be about 75mm to the side of the bike centreline which in itself will cause the vector of the combined forces to try and stand the bike back up. This will require a slightly greater lean angle to counteract.

If the resultant vector acted through the point of contact of the tyre when the bike was upright your formulae would be correct but they aren't taking into account that the contact point is actually a curved tyre and the point at which it contacts the road is displaced progressively to the side as the bike is leaned. The resultant vector still has to act through the centre of mass and the contact point of the tyre. The vector angle will stay the same but the bike will be leaning further because of the offset contact point of the tyre. I think!

The question was about "bike" lean angle, not "vector" angle.

[avgas]

Rotational G???

Yep - ever rode a bike with big heavy swinging round bits, makes the corner a whole lot more interesting.

But alas you all still haven't got the joke, constant speed, angle with 0.5G!!!

[cooneyr]

Hmmm, I must say I'm not too happy about how you put it. There's a problem with causality!......

Correct if you are looking at it from the perspective of how fast can I take a corner without falling off. This whole thread has been a bit backwards though, and we are looking at it from the perspective of how do I achieve horizontal acceleration vector of 0.5.


.......The question was about "bike" lean angle, not "vector" angle.

We didn't get a bike specified but if you are going to get that pedantic then the whole question is a bit farcical really. This is why I have stuck with resultant vector angles rather than bike angles. They are two very different things and unless we have details of the bike itself, the type of riding we are talking about (remember dirt/supermoto riders do it different to road riders) and the bike to rider relationship.

Lets go back to the pole's level of sensitivity. I'm guessing the answer is probably "somewhere around 20 degrees" for a road rider but I don't really know.

Cheers R

[GSVR]

Whats interesting is the spread of the pollsters.

Wonder what the spread would have been if the question was posed as to the maxium G force achievable on the bikes they ride. So many variables such as clearance, width of bike and rider and grip of tyres.

Wonder how many would say they pull more than 1G. And how many would say over 1.5G.

[avgas]

Wonder what the spread would have been if the question was posed as to the maxium G force achievable on the bikes they ride. So many variables such as clearance, width of bike and rider and grip of tyres.
Exactly, hence the joke.
Say you pull 0.5G in corner on a moto-gp bike, then do the same on the munch mammut.
They tyres of the mammut would find it incredibly hard lean angles aside. Throw constant speed in there and its like rubbing the tyres down with vaseline.
I have never done a constant speed corner. I pray to the gods i never have too. May work on the new 2WD bikes though. depending on momentum balances.

[*caution*]

I have never done a constant speed corner.

you've never gone round a corner without changing speed? or am i missing something? most likely, its friday arvo....

[speedpro]

"AVGAS" lost me there. Surely if .5G is one vectors worth of force and gravity is the other vectors worth of force the resultant combined vector will be the same for both bikes, and not that terribly difficult to achieve either.

Humour me, I'm not an engineer.

[Badjelly]

Yep - ever rode a bike with big heavy swinging round bits.
Wheels?


But alas you all still haven't got the joke, constant speed, angle with 0.5G!!!
No.

Excuse me while I ride around Cobham Drive roundabout at, say, 50 km/h, about 25 degrees lean, 0.5 G. I think I can keep the big heavy swinging round bits from going USD.

[*caution*]

But alas you all still haven't got the joke, constant speed, angle with 0.5G!!!

Is the joke your trying to get at, that its constant speed, but its accelerating (0.5G)? because I figure hes implying constant speed perpindicular to the radius of the bend..... while of course the bike is also accelerating toward the centre of the bend (basic rotational mechanics) or else it wouldn't be turning.

enough for a friday anyway....

[Mikkel]

But alas you all still haven't got the joke, constant speed, angle with 0.5G!!!

I think it would be worthwhile to point out that while velocity is a vector (it has a magnitude AND a direction) - speed is just the magnitude of the velocity vector.

You can have acceleration change the direction of the velocity vector - but not the magnitude - thus maintaining a constant speed.

[McJim]

What a silly place to ask this question. You're troll aren't you?

I just go round corners and lean just enough not to fall off.:rofl:I tend to leave my protractor at home.

[speedpro]

I've hopped off my bike mid-corner a few times. Alas I didn't think of carrying my protractor so I could measure what was happening. Some of it was pretty obvious though, the old sky,earth,sky,earth thing and once there was an "F5 Dave" in there somewhere as well.

[Cruisin' Craig]

O.K then. Take a look at the attached picture if you will.

The horizontal line at the bottom is the ground, and the diagonal line represents the bike at some angle of lean.

AC represents the downward acceleration of 1.0G, which is equal to the two component vectors CB and BA.

If you take a moment to check, you will see that x is equal to the angle of lean.

Now if you look at the first triangle: ABC, you can see that it has three angles. One is 90 degrees, one is x, and the other is 90-x degrees.

Look now at the smaller triangle BCD, and you will see that one angle is also 90 degrees, one is also x, and the other must therefore also be 90-x degrees. This means that ABC is exactly the same shape as BCD, just bigger.

This means that the ratio between AC and BC (the hypotenuse of each triangle) is equal to the ratio between BC and BD (the opposite of each triangle). Or in other words: AC/BC = BC/BD.

Multiply both sides by BC*BD and we get AC*BD = BC*BC

Note also that CD = 0.5G as this represents the centripetal acceleration.
This means that AC (also known as G) = 2*CD. Substitute this into the above equation and we get BC*BC = 2*CD*BD

Then we use pythagoras on the smaller triangle to get:
BC*BC=BD*BD+CD*CD

Or substituting BC*BC with the equation from two lines up , and you are left with:
BD*BD+CD*CD=2*CD*BD

which can be re-written: BD*BD - 2*CD*BD + CD*CD = 0

or Factorising: (BD-BC)*(BD-BC)=0

This shows that BD = CD, and if you look at the diagram this clearly means that x = 45 degrees.

The 26 or so degrees spoken of earlier was correctly calculated, but this is the angle of the total accelerative force, and is not the same as the angle of lean (as has already been said).

The angle of lean is 45 degrees.

[ital916]

Thats what I was saying, for 5khr, slow, tight turn, small radius to create 0.5G.

Can't say i've got my book on me, I remember we used engineering mechanics: dynamics, (meriam and kraige if i rememeber correctly)!! Vectors are king, who was that lecturer agian, simon B....someone i think, he had this slide he would put up "Only a fool doesn't draw a vector diagram!" haha

Haha gotta love the dynamics text book. Guess we will be wrapping our heads around this kind of problem this year in eng. :whocares:....I care..cos engineering is the "shizzle", especially when people do awesomely useless stuff, just cos they can. Like the Y2K.

[ital916]

O.K then. Take a look at the attached picture if you will.

The horizontal line at the bottom is the ground, and the diagonal line represents the bike at some angle of lean.

AC represents the downward acceleration of 1.0G, which is equal to the two component vectors CB and BA.

If you take a moment to check, you will see that x is equal to the angle of lean.

Now if you look at the first triangle: ABC, you can see that it has three angles. One is 90 degrees, one is x, and the other is 90-x degrees.

Look now at the smaller triangle BCD, and you will see that one angle is also 90 degrees, one is also x, and the other must therefore also be 90-x degrees. This means that ABC is exactly the same shape as BCD, just bigger.

This means that the ratio between AC and BC (the hypotenuse of each triangle) is equal to the ratio between BC and BD (the opposite of each triangle). Or in other words: AC/BC = BC/BD.

Multiply both sides by BC*BD and we get AC*BD = BC*BC

Note also that CD = 0.5G as this represents the centripetal acceleration.
This means that AC (also known as G) = 2*CD. Substitute this into the above equation and we get BC*BC = 2*CD*BD

Then we use pythagoras on the smaller triangle to get:
BC*BC=BD*BD+CD*CD

Or substituting BC*BC with the equation from two lines up , and you are left with:
BD*BD+CD*CD=2*CD*BD

which can be re-written: BD*BD - 2*CD*BD + CD*CD = 0

or Factorising: (BD-BC)*(BD-BC)=0

This shows that BD = CD, and if you look at the diagram this clearly means that x = 45 degrees.

The 26 or so degrees spoken of earlier was correctly calculated, but this is the angle of the total accelerative force, and is not the same as the angle of lean (as has already been said).

The angle of lean is 45 degrees.

I find with mathematics, if you try hard enough you can prove that something is something else using enough mathematical know how. A prim example is of a maths teacher proving 2 = 3. Gotta find it on google.

[DingoZ]

Get up cause it's so :sunny:..
Yes thinks it's :2thumbsup: Great day for a ride
Go out and start the Bike
:yes: I'm gunna enjoy this ride, cause I learn something every time
:shit: Got that corner a little bit wrong
:woohoo: Got the Groove now
Fark I like this motorcycling stuff - why did I wait so long

:eek5: The above stuff has got me all :confused:

:Oi:.....Everyone :Playnice: and let's all enjoy the :ride:

[Cruisin' Craig]

I find with mathematics, if you try hard enough you can prove that something is something else using enough mathematical know how. A prim example is of a maths teacher proving 2 = 3. Gotta find it on google.

It won't be a real proof though. If you find it and post it, I'm willing to bet I can show where they cheated :-)

[ital916]

I'll try my darndest to find it. I wonder if we will build anything cool in the third year eng :whistle:..jet bike anyone lol

[Disco Dan]

Five pages and no G-spot... not even a G-string... :weep::weep:

[Jantar]

O.K then. Take a look at the attached picture if you will.

The horizontal line at the bottom is the ground, and the diagonal line represents the bike at some angle of lean.

AC represents the downward acceleration of 1.0G, which is equal to the two component vectors CB and BA....

And that is where your analysis fails. AC is NOT the resultant, it is a component vector. CD is the other component, and AD (not drawn) is the resultant. Now try again.

[Cruisin' Craig]

And that is where your analysis fails. AC is NOT the resultant, it is a component vector. CD is the other component, and AD (not drawn) is the resultant. Now try again.

I have broken AC down into two vectors. Therefore those two vectors are the component vectors of AC, and AC is the resultant of those two vectors. Get it?

Now if you want to add two different vectors together to get a different resultant vector that's fine, but it doesn't affect the truth or otherwise of my argument.

[EnzoYug]

Does the corner tighten up just before you realise you've gone in too hot?
Is there loose gravel on the road?

I've run a few tests and found 90' to be about right - it depends if you're bouncing alongside or underneath the bike as it grinds away the plasticy bits.

[Jantar]

I have broken AC down into two vectors. Therefore those two vectors are the component vectors of AC, and AC is the resultant of those two vectors. Get it?

Now if you want to add two different vectors together to get a different resultant vector that's fine, but it doesn't affect the truth or otherwise of my argument.

AC is a vector. It has direction = towards the Earth's center, and it has magnitude = 1.0 G.

Your diagram gives AB at direction x as a component of AC, however it is the resultant of AC and is the hypotenuese. ACB is the right angle, not ABC.

[Cruisin' Craig]

.....ACB is the right angle, not ABC.



AC is directed straight down, and CB represents the bike at some angle of lean. So ACB can't be a right angle unless the bike is horizontal.

[Jantar]

... So ACB can't be a right angle unless the bike is horizontal.
Of course it can. ACB = 90, ABC = 90-x, and BAC = x.

Also don't forget that the sum of All vectors must add to 0, otherwise the entire system will be unbalanced.

[Cruisin' Craig]

Of course it can. ACB = 90, ABC = 90-x, and BAC = x.

Also don't forget that the sum of All vectors must add to 0, otherwise the entire system will be unbalanced.

Ah, I see. Quite right, ACB can be the 90 degree angle. If you do it that way the calculation ends up just the same except we have the bike leaning the other way.

I've also just done another calculation involving balancing torque moments and still get 45 degrees. I'll post that a bit later tonight once I've had a chance to check it to see what people think. Turns out it is a far easier calculation to do as well.

I'm still standing by my figure of 45 degrees, but I'm happy to reconsider if you can prove that the resultant acceleration vector being 26 degrees from vertical means that the lean angle must also be 26 degrees, as I believe they are not necessarily the same.

[cooneyr]

I think you might be a little confused with all that analysis Cruisin' Craig. Like Jantar said AC is a component as is CD and the resultant vector is AD. Gravity is towards the earth's centre hence becomes one of the fixed axis of the system. We were also given that the 0.5 G is horizontal (meaning it is at 90 degrees to AC) hence the other fixed axis.

Think about what we actually know and are trying to work out not what we think we know and working backwards. Also dont worry about the algebra - we learn trig to make life easier and trig is less prone to errors.

Cheers R

[speedpro]

I've also just done another calculation involving balancing torque moments and still get 45 degrees..

Without being a genius, 45 deg would be obtained if both the vertical (gravity - 1G) vector and cornering forces vector were equal and therefore 1G as well. The resultant vector would be at 45deg and have a value of square root of 2. Simple trigonometry.

[Mikkel]

Also don't forget that the sum of All vectors must add to 0, otherwise the entire system will be unbalanced.

Well, you don't want the sum of the vectors to be zero if you're trying to go around a corner now... ;)

[madandy]

This thread has enlightened me somewhat regarding the constant fuck ups Civil Engineers make .

[cooneyr]

This thread has enlightened me somewhat regarding the constant fuck ups Civil Engineers make .

Like this you mean? :D

Cheers R

[madandy]

That's exactly right! Now they stand there for days 'discussing' how they fucked up so bad.
I have worked with mullet haired, jail tatt'd, toothless, chain smoking, compulsive cursing machine operators who make more sense and end up correcting the balls up's thier 'superior' Engineer supervisors create...
Now they argue about lean angles and come up with seemingly infinte reasons why they cannot agree.
Dont get me started on 'mechanical engineers'...

[dipshit]

This thread has enlightened me somewhat regarding the constant fuck ups Civil Engineers make .

haha... yes, they sit around with their calculators and slide rulers, but half the time haven't got a clue about reality in the real world.

[GSVR]

I so like Google

http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=312

If your've got Microsoft Excel or the veiwer this is very good

http://www.msgroup.org/images/bike.xls

27 degrees is close enough.

[cooneyr]

..... I have worked with mullet haired, jail tatt'd, toothless, chain smoking, compulsive cursing machine operators who make more sense and end up correcting the balls up's thier 'superior' Engineer supervisors create......

Thats cause engineers never have perfect knowledge of a site and are always accountable to the accountants. Its easy to bitch about a design with the benefit of more detailed info - i.e. the hole in the ground that shows the subgrade is rubbish. You are right though most of those old toothless guys are bloody good to have on site - as long as you can work with them and not argue every 2 seconds.

Ever heard the saying "An engineer can do for $0.02 what any fool can do for $2."


I so like Google

http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=312

If your've got Microsoft Excel or the veiwer this is very good

http://www.msgroup.org/images/bike.xls

27 degrees is close enough.

And I though we were being geeky! That xls file is a bit excessive! :D

Cheers R

[speedpro]

Yeah like the engineer who specified square roof penetrations on a job I was doing. My suggestion of specifying round holes so we could use a holesaw to make them hadn't been thought of. There were a few other "issues" as well. Me and a mate used to regularly pick holes in an engineer mates designs for all sorts of things, mainly his self designed sports car rear-end. ZZR1100 engined so kind of relevent on a KB thread. maybe. He just about always came round to our idea, after say all the teeth were ripped off both sprockets.

[Jantar]

... Dont get me started on 'mechanical engineers'...

Well you do realise that mechanical engineers build weapons, and civil engineers build targets.

[madandy]

Ever heard the saying "An engineer can do for $0.02 what any fool can do for $2."
Cheers R


Nope. I just watch them blow budgets by a few million because they fail to grasp sufficient knowledge of a site.
Theory is only half of it.

[Cruisin' Craig]

O.K then. I've now looked at several sites that do the calculation in a couple of different ways, and I grumpily concede defeat. :angry:

[FROSTY]

WHERE is the box that says --I lean far enough to get round the bloody corner

[Mikkel]

That's exactly right! Now they stand there for days 'discussing' how they fucked up so bad.
I have worked with mullet haired, jail tatt'd, toothless, chain smoking, compulsive cursing machine operators who make more sense and end up correcting the balls up's thier 'superior' Engineer supervisors create...
Now they argue about lean angles and come up with seemingly infinte reasons why they cannot agree.
Dont get me started on 'mechanical engineers'...

Well, cock-ups happen. But on the other hand I'd like to see Joe average try and design a two kilometer long bridge or tunnel, built/dug from each end and have them meet up at the middle.

Truth be told they get it right almost every time.

Also true that an engineer who doesn't listen to the guy who's going to do the work is a fool beyond hope. Planning is one thing - execution another.

However, look for a second at what the planning, design and knowledge that is engineering has made us capable of. How'd you like to catch a ferry to cross Auckland harbour - oh, forget that, the ferry is engineered as well. How about having to slosh through your own and everybody elses shit because no one ever thought about a sanitation network... No, be happy that there are people out there who does the civil engineering bits.

And sure, these days budgeting is atrocious. However, I wouldn't blame the civil engineers. They are most likely told by someone further up the ladder (no Dan, not that ladder) that we want to build a bridge here or there and it can't cost more than this or that - make it happen or look for another job.
Oh the examples of creative budgeting and planning I could relate... Usually it's the bloody bureacrats who instigate it.

[madandy]

Ya took a while :)
http://i19.photobucket.com/albums/b160/maddandy/istockphoto_440872_hook_line_and_si.jpg

[avgas]

You can have acceleration change the direction of the velocity vector - but not the magnitude - thus maintaining a constant speed.
So acceleration doesn't change the magnitude of speed? May i ask what does :) You are partially right - but you must see the joke i have here.

[Mikkel]

Ya took a while :)
http://i19.photobucket.com/albums/b160/maddandy/istockphoto_440872_hook_line_and_si.jpg

Always happy to bite. ;)


So acceleration doesn't change the magnitude of speed? May i ask what does :) You are partially right - but you must see the joke i have here.

I think I see what you mean. Speed being a scalar hasn't got a magnitude - only a value. That value being the magnitude of the velocity...

Edit: A joke people taking an interest in this thread might appreciate: http://www.xkcd.com/123/

[avgas]

I think I see what you mean. Speed being a scalar hasn't got a magnitude - only a value. That value being the magnitude of the velocity...
Don't forget while speed may be considered a scalar, integration/diff brings you back and forth between Acceleration/Speed/Distance with respect to time.
While for theory's sake you could say that time is nothing. This actually brings all components to zero.
Likewise if no distance is covered it than is there an acceleration?
If there is no acceleration? is there force?
Which brings me to case and point of the joke, without knowing the distance covered over a period of time (i.e. the speed) or considering this constant (easiest is to consider 1 - as multiples of 1 do not affect equations) what angle do you feel 0.5G.
Well i think if you were to lean around 45 degrees you may feel that - because at that point it would be your leg hold half the weight of bike off the road :)
Now do you see the joke, its nothing to do with the angle more the speed and the corner distance. The angle helps you compensate for the G force. Motards/Sidecars dont lean do they - they slide. Do they experience 0G?

[Badjelly]

You certainly like to make things complicated, don't you avgas?

[avgas]

You certainly like to make things complicated, don't you avgas?
The only simply G force is the one you feel on the earths surface. And even that one seems to f people up from time to time. But yes your are correct calculating G-Forces on a bike is like deriving a PID controller for a Space shuttles pitch.

[Mikkel]

Don't forget while speed may be considered a scalar, integration/diff brings you back and forth between Acceleration/Speed/Distance with respect to time.
While for theory's sake you could say that time is nothing. This actually brings all components to zero.
Likewise if no distance is covered it than is there an acceleration?
If there is no acceleration? is there force?
Which brings me to case and point of the joke, without knowing the distance covered over a period of time (i.e. the speed) or considering this constant (easiest is to consider 1 - as multiples of 1 do not affect equations) what angle do you feel 0.5G.
Well i think if you were to lean around 45 degrees you may feel that - because at that point it would be your leg hold half the weight of bike off the road :)
Now do you see the joke, its nothing to do with the angle more the speed and the corner distance. The angle helps you compensate for the G force. Motards/Sidecars dont lean do they - they slide. Do they experience 0G?

Nope - I'm just very very confused now.

[avgas]

Nope - I'm just very very confused now.
Ok ok, i'm getting ahead of myself a bit mabey.
Go out on the bike, do a constant 1kph round a corner at 26 degree lean angle. How many g do you feel?
Now do the same corner at 100kph round a corner at 26....? how many g do you feel?
Now go out at 0kph in the same experiment? at what angle to do experience 0.5G?

[cooneyr]

Ok ok, i'm getting ahead of myself a bit mabey.
Go out on the bike, do a constant 1kph round a corner at 26 degree lean angle. How many g do you feel?
Now do the same corner at 100kph round a corner at 26....? how many g do you feel?
Now go out at 0kph in the same experiment? at what angle to do experience 0.5G?

Yerbut - that wasnt the question, your starting at the wrong point. There is defiantly a speed and radius component required to achieve a 0.5G horizontal acceleration. The point of this question though is assuming you are pulling 0.5G (at whatever speed that takes on a given radius) how much would be leaning. The lean angle doesnt change for a a given horizontal acceleration.

If you vary the speed for a given radius the horizontal acceleration changes or if you vary the radius for a given speed the horizontal acceleration changes.

Coming back to your side car point. The side car doesnt lean (well it does a little bit) but the angle of the resultant vector (result of 0.5G horizontal and 1G vertical) is still at 26.5 degrees. For a two wheeled motorbike it must lean else the horizontal acceleration makes you end up on your arse.

Cheers R

[Mikkel]

Ok ok, i'm getting ahead of myself a bit mabey.
Go out on the bike, do a constant 1kph round a corner at 26 degree lean angle. How many g do you feel?
Now do the same corner at 100kph round a corner at 26....? how many g do you feel?
Now go out at 0kph in the same experiment? at what angle to do experience 0.5G?

cooneyr did a good job. :niceone:
One thing that's worth mentioning though is that it's not the lean angle of the bike - it's the lean angle of the centre of mass. I.e. what is the angle between vertical and the vector that goes from the contact patch through the centre of mass. Which is why you can corner sharper by hanging off.
Obviously you can't do that in a car - the result here is that there's more force on the outside wheels than the inside to reflect this. This is one of the main reasons for desiring a centre of mass as low to the ground as possible in cars since the higher up it is - the bigger the difference between inside and outside wheels.

However, I do believe most of this was covered earlier in the thread. :yes: