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Topic: "The Physics Problem" (Read 4 times) previous topic - next topic

NI$HANT

If you see the following video >> BMW S1000RR CRASH - http://youtube.com/watch?v=_A_PFBRAx9w

You will see that the riders AT A CERTAIN SPEED , WITH A GIVEN WEIGHT , ALONG WITH A CERTAIN ANGLE OF TILT goes down as the gravitational pull acts like that(may be I don't know better)

I actually wanted to know that one can know using physics that he can calculate that how much he can successfully tilt the bike like in the video GIVEN (he knows the weight of the vehicle in total (including him) + knows the Gravitational pull + the speed )

Assumptions can be:
a) The road is not wet.
b) Normal sunshine is there making the road a normally hot
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Boffin1

His back wheel lost traction and slid out, its not very mathematical as rubber " marbles" , oil, or a damp patch, can cause it.

You only find where the limit is when you have hit it, which can be expensive   :-(
With my mobile phone I can call people and talk to them -  how smart can you get ?

sbright33

There is a formula with 3 variables and a constant.  They are speed, radius, and lean angle.  The constant only depends on the tires, bike, rider geometry/mass.
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Mad physicist

#3
Jul 11, 2013, 06:29 pm Last Edit: Jul 11, 2013, 08:01 pm by Mad physicist Reason: 1
There is a typical undergrad problem involving an inclinated wheel travelling on a circle.

I did the quick back-of-the-envelope travel agency paper calculation (attached). On this extremely simplified example, you basically have to equate two torques : one from the inclination of the wheel (which tends to make it fall) and one from the change of wheel rotation vector due to orbital movement (which tends to set it back upright). (I neglected precession -due to the rotating wheel inclination- as wheels on a bike are much more constrained; the front one is free but controlled by the driver and the rear one is fixed. I didn't checked if the effect is actually negligible, but it seems an honest approximation, espically for a bike.) Note that for this example, the normal and friction forces can be made irrelevant by choosing the origin of axes at the contact point, but these simplifications most probably won't be possible when you fully consider the bike.

I probably made some calculation mistakes (i make mistakes virtually all the time), but the answer for the equilibrum lean angle seems, at least, reasonable. It only depens on the whell mass and radius, its speed and the radius of the big circle it travels (i left the rotational speed versions of these, but it's the same thing).

What changes for a full bike ? Obviously the center of mass vector won't be as simple (depends on the bike mass distribution) ; the inertia moment (but i left I anyway) ; and finally the radius R is to be changed with the curvature radius of the bike's path (which is now a function of theta) ; plus some other things (like neglected friction or what the driver does with the front wheel).

But to answer your initial question, yes it is possible even for a real bike, and solving everyting numerically should be easier in this case.


edit: On the lowest line, i wrote dr/d(theta), it's of course d(u_r)/d(theta), the unit vector. But that's what i meant on the next step.

NI$HANT

Thanks all specially Mad physicist I check it out deeply in sometime
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NI$HANT

@ Madphysicist Can you make it more clear in writing please!
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Mad physicist

#6
Jul 12, 2013, 02:12 am Last Edit: Jul 12, 2013, 02:16 am by Mad physicist Reason: 1
Yeah sorry, that was really back-of-the-envelope. I'm typing a PDF version right now, aslo corrected from the mistakes I previously made (including quite stupidly calling theta both the coordinate and the lean angle :smiley-roll-sweat:). It'll be ready soon.

Mad physicist

#7
Jul 12, 2013, 03:15 am Last Edit: Jul 12, 2013, 01:24 pm by Mad physicist Reason: 1
Hi,

While typing, I found this wesbite, which presents a simpler way of doing it : http://www.real-world-physics-problems.com/bicycle-physics.html, but they say they have "neglected 3D effects". My answer is the same as their (given that R*Omega^2 = v^2/R), if you neglect the 2*R*I*Omega*omega term.

Anyway, here is the PDF. I actually made two sign mistakes in my first thing, but they cancelled each other! How lucky, why doesn't this happen at exams ?

NI$HANT


hehe, Sometimes teachers while checking offer discount.

what I now want is that the gravitational(gravity Sensor) quantities & Speed derived from a electronic circuitry of mine along with Constants fed into the Final Equation and get the angle,all this need to be calculated in c/c++ with available moth functions library, how to go about that?
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cjdelphi

I'm amazed there's no gyroscope/accelerometer to give warnings to the rider that they've about to hit point of no return...

some kind of buzzer to alert the rider....

NI$HANT

@ cjdelphi

The gyro/ accelerometer is needed to note the angle Not a Buzzer at all as these Super Sport or sports Bikes make a lot of noise So most Probably an LCD indication.
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Chagrin

I think you're going to get a lot of error in your calculation due to the leaning tire and how its grip of the road will change. With the bike and rider you've got ~700 lbs of force (guessing) on a ~40 PSI tire and a footprint of (700/40) 17.5 square inches. But because you're sitting more on the sidewall of the spinning tire that rubber footprint is moving at different speeds across its width and gripping the road much differently than if the bike was upright.

NI$HANT

A good point made here chagrin but the Constants Such as the tyre Radius when Bike is upright & when the Bike leans Can be very easily measured and input into a variable & the acuter met thee accelerometer can easily make out that How much the Bike is leaning So as to Consider that tyre's Radius along its width.
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GoForSmoke

Mad Scientist, hope you remember to put centripetal force in there? It's a biggie on a 500+ lb bike cornering at high speed.

Quote

I'm amazed there's no gyroscope/accelerometer to give warnings to the rider that they've about to hit point of no return...

some kind of buzzer to alert the rider....


Usually the point where you have to lift the inside foot because the peg is scraping provides a clue that that's all the lean you're going to get. On my bikes I could lean less to the right because the pipes scraped a bit sooner.

Human gyro/accelerometer is feeling where the G's are pushing. It should be directly into the seat but you can shift it a bit to get the bike more upright if you can lean yourself more into the turn which is hanging over the line just a bit btw.

The accident was determined by the entry speed to the turn. Warnings after that just let you choose to go down or leave the road on the outside of the turn.


I find it harder to express logic in English than in Code.
Sometimes an example says more than many times as many words.

Mad physicist

#14
Jul 12, 2013, 09:07 pm Last Edit: Jul 12, 2013, 09:14 pm by Mad physicist Reason: 1
Quote

Mad Scientist, hope you remember to put centripetal force in there? It's a biggie on a 500+ lb bike cornering at high speed.

Centripetal force is there indeed. What I didn't include was friction and normal force. For a thin wheel, one can approximate they lie on the origin of axes (as i draw them), which is exactly crossed by th wheel longitudinal axis (hence they don't enter into the moment calculation). But for a broad wheel like there are on motorbikes, the point of application will be obviously offset. This is not difficult to include conceptually, but that gores up (does that word exist?) the formulas a bit.

Note that, with my formula, phi always has a solution, which means the bike wheel can be at equilibrium at any lean angle. This is obviously not true in reality (the bike falls), where more stuff (see previous post) must be taken into account.

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