riding physics.

[ADoMinGu3z]

so i kno that this forum is home to many different types of people and was hoping if there was anyone out there that could explain to me the physics of riding. such as normal forces, frictions forces, maximum velocities in a turn and how lean allows the bike to turn at higher speeds. also how to calculate the maximum lean angle in certain situations. id ask my physics professor, but i thought id try here first

[Motofool]

Andrew,

I would love teaching you what I know about the subject.

Start by reading these threads:

http://www.ninjette.org/forums/showthread.php?t=121404

http://www.ninjette.org/forums/showthread.php?t=119958

http://www.ninjette.org/forums/showthread.php?t=114555

http://www.ninjette.org/forums/showthread.php?t=114372

http://www.ninjette.org/forums/showthread.php?t=100964

This is a great book:

http://books.google.com/books?id=rJT...page&q&f=false

[ADoMinGu3z]

thanks. last link was gold

[Motofool]

You are welcome

Ask all the specific questions that you want.

[GearMonkey]

Watch A Twist of the Wrist 2, by Keith Code.
This may not go in as much physics depth as you're looking for, but it's a good learning video.

[ADoMinGu3z]

in the event that the road itself is at an angle, say 3 degrees, is there any way to calculate a maximum velocity one can maintain in a curve without washing out and . i know its all relative to friction, weight, lean angle and other conditions, but say all those could be calculated. in other words, the theoretically perfect turn.

[Motofool]

Yes, it is very simple.

The tangent of that angle equals (V^2*/gR)

Being V the speed of the bike, g the gravity acceleration and R the radius of the turn.

We assume that any decent street tire on dry pavement can develop a coefficient of friction around 1.0

You multiply that number (or a lower one for wet condition, concrete, etc.) by the percentage of the weight of the bike and rider on the specific contact patch (ideally is 40/60 under acceleration).

That is the maximum lateral friction that you have for that patch.

[ADoMinGu3z]

thanks, how would the camber of the road affect the lean angle and speed. id assume you would b able to go faster because the normal force would b more perpendicular to the road itself.


if you dont mind me asking...what do you do for a living haha

[choneofakind]

Welcome Andrew.

Good to see that you've met Hernan, our resident physics buff

Jiggles just posted up a video thread designated for nerdy science videos. Feel free to peruse that one as well. It's got some fun stuff in it, like this video:

Link to original page on YouTube.

[ADoMinGu3z]

Originally Posted by choneofakind

Good to see that you've met Hernan, our resident physics buff

This made me laugh. im pretty sure that although the physics behind this isnt hard, my professor would have had no idea what i was talking about. shes a russian astrophysicist with a heavy accent. glad to see someone with the same passion, willing to break it down mathematically

ISNT MATH FUN!

[choneofakind]

The normal is a vector that's always perpendicular to a surface. There's really no more or less perpendicular about it. It's always 90 degrees

Assume 2 turns of equal radius. One is banked, and the other is off-camber. You take both at the same forward speed.

In the off-camber turn, your lean angle in reference to the normal force of the road will be larger, meaning your tire thinks you're leaning futher. For the turn that's banked, your lean angle in reference to the normal force of the road will be smaller, meaning your tire thinks you're leaning less. However, your angle with the horizon will still be the same in both situations because your speed is still the same, requiring the same centripetal force to turn.

[ADoMinGu3z]

so the angle is the same relative to the reference frame, its just the reference frames that are different correct?

[choneofakind]

The overall lean angle is the same, but the angle in reference to the ground will be different. It all depends what perspective you're looking at it from.

But just to be clear, things drag quite easily on the ninjette. Many of us have already proved this. Don't go testing how far you can lean on a turn just because you can calculate how fast you can take it.

[Motofool]

Originally Posted by ADoMinGu3z

thanks, how would the camber of the road affect the lean angle and speed. id assume you would b able to go faster because the normal force would b more perpendicular to the road itself.


if you dont mind me asking...what do you do for a living haha

Read about road banking in the attachment.

The camber of the road has no effect on the lean angle and associated speed.
If that camber is positive, it reduces the lateral friction forces on the contact patches.
If that camber is negative, it increases those forces.

The normal force is a reaction force that is always perpendicular to the action of the weight exerted perpendicularly to the road surface.

I am semi-retired now, but I used to do similar calculations for some time.

[CC Cowboy]

Andrew, I hope you get an "A" on the test.

[oblivion007]

Ugh lol the physics of riding can be very difficult analytically. As for the questions you asked they're all pretty relatively easy to answer within a boundary such as taking into consideration a wheel that isn't perfectly balanced. No time now so maybe I can answer the questions later.

[oblivion007]

Originally Posted by Motofool

Read about road banking in the attachment.

The camber of the road has no effect on the lean angle and associated speed.
If that camber is positive, it reduces the lateral friction forces on the contact patches.
If that camber is negative, it increases those forces.

The normal force is a reaction force that is always perpendicular to the action of the weight exerted perpendicularly to the road surface.

I am semi-retired now, but I used to do similar calculations for some time.

If by camber you mean the road is banked then it does have an effect on the maximum lean angle relative to the horizontal (not parallel to the road but to perpendicular to the direction of gravity in the local area) and the maximum velocity around the corner because it shifts more of the load from the frictional force (static) to the normal force where the normal force to my understanding has a much higher threshold for failure.

Its like comparing sliding your tires while leaned over to how much weight you can put on your bike standing up without failure.

[Motofool]

Originally Posted by ADoMinGu3z

This made me laugh.......

ISNT MATH FUN!

Actually, Chris (Chonefakind) is currently studying all this and much more complex subjects at the University; he knows what he discusses.

Now, going back to the previous formula:

Since we have assumed 1.0 for the coefficient of friction and tg 45 = 1.0, for a bank of 45 degrees, which is around the maximum lean angle for a stock Ninjette, the lateral forces will equal the gravitational forces on the contact patches.

Hence, you could calculate the theoretical maximum speed for each turn the following way:

Vmax = 0.68 * sqrt (Rg)

For example, I want to calculate the top speed for a roundabout that I ride every morning.
I went to Google Earth and measured the radius, being 200 ft.

Vmax = sqrt (200 ft * 32.2 ft/s^2) = sqrt (6440) = 80 ft/s

mph = 0.68 ft/s

Vmax = 0.68 * 80 = 54 mph

Speed limit for that roundabout is 35 mph.
There is no legal speed on the USA's streets that will allow you to lean your bike more than around 30 degrees.

Note that we are referring to the lean angle formed by the system's center of gravity (or combined CG) against the vertical.

Depending on a rider hanging off or not and the width of the tires, the lean angle of the bike will be different.
However, the forces on the contact patches depend only on the angle of the CG.

The advantage of keeping the bike more straight up is that the suspension will work better regarding keeping the rubber patches in contact with the pavement.

[Motofool]

Originally Posted by oblivion007

If by camber you mean the road is banked then it does have an effect on the maximum lean angle relative to the horizontal (not parallel to the road but to perpendicular to the direction of gravity in the local area) and the maximum velocity around the corner because it shifts more of the load from the frictional force (static) to the normal force where the normal force to my understanding has a much higher threshold for failure.

Its like comparing sliding your tires while leaned over to how much weight you can put on your bike standing up without failure.

I wrongly used the word camber, but it should be cant or superelevation:

http://en.wikipedia.org/wiki/Cant_(road/rail)

I cannot understand your statement completely.
There is a unique force acting between the road and the tire; however, we like make it sub-forces aligned in a way that we can better understand.

Those bikes riding inside that mesh sphere at the circus can turn in any plane, if they develop enough speed, since the surface is always perpendicular to the tires; hence, their lean angle is always zero.

[oblivion007]

Originally Posted by Motofool

I wrongly used the word camber, but it should be cant or superelevation:

http://en.wikipedia.org/wiki/Cant_(road/rail)

I cannot understand your statement completely.
There is a unique force acting between the road and the tire; however, we like make it sub-forces aligned in a way that we can better understand.

Those bikes riding inside that mesh sphere at the circus can turn in any plane, if they develop enough speed, since the surface is always perpendicular to the tires; hence, their lean angle is always zero.

Yes relative to the surface the tires make contact with. I don't know what the community thinks of the term 'lean angle' but I take it to be the angle between the vertical (direction of gravity) as opposed to the normal of the surface. Anyways I learned some new terms from the pdf you posted lol, thanks.

[jpbear]

Originally Posted by Motofool

Actually, Chris (Chonefakind) is currently studying all this and much more complex subjects at the University; he knows what he discusses.

Now, going back to the previous formula:

Since we have assumed 1.0 for the coefficient of friction and tg 45 = 1.0, for a bank of 45 degrees, which is around the maximum lean angle for a stock Ninjette, the lateral forces will equal the gravitational forces on the contact patches.

Hence, you could calculate the theoretical maximum speed for each turn the following way:

Vmax = 0.68 * sqrt (Rg)

For example, I want to calculate the top speed for a roundabout that I ride every morning.
I went to Google Earth and measured the radius, being 200 ft.

Vmax = sqrt (200 ft * 32.2 ft/s^2) = sqrt (6440) = 80 ft/s

mph = 0.68 ft/s

Vmax = 0.68 * 80 = 54 mph

Speed limit for that roundabout is 35 mph.
There is no legal speed on the USA's streets that will allow you to lean your bike more than around 30 degrees.

Note that we are referring to the lean angle formed by the system's center of gravity (or combined CG) against the vertical.

Depending on a rider hanging off or not and the width of the tires, the lean angle of the bike will be different.
However, the forces on the contact patches depend only on the angle of the CG.

The advantage of keeping the bike more straight up is that the suspension will work better regarding keeping the rubber patches in contact with the pavement.

Lovely, the roundabout near my house has a diameter of 213 feet - radius 106. the formula gives me the max speed as 39.somethingmiles - roundup to 40mph. good to know..

[Motofool]

Originally Posted by jpbear

Lovely, the roundabout near my house has a diameter of 213 feet - radius 106. the formula gives me the max speed as 39.somethingmiles - roundup to 40mph. good to know..

You should not be riding on public streets at speeds close to that theoretical max.

Oil or water on pavement (even when not visible), surface irregularities, tire's pressure and temperature, mid-turn steering corrections among other factors may made you lowside.

It is always wise keeping a safety margin regarding speed while sharing the street with others.

Top speeds are only for the track, which is a safer environment due to the absence of dangerous cars and obstacles.

This thread is discussing just the physics of riding; practical street riding is a different subject.

[blink]

This is why I gave up on physics...

[Motofool]

Originally Posted by blink

This is why I gave up on physics...

Why?

If you ride, you didn't give up, you use it each time you take your bike out.

You refill oil and lube your chain to reduce friction, you select your tires and brake pads to increase friction, you balance your CG over the contact patches, you counter-steer, you give gas to increase acceleration and by switching gears, you are applying the old lever principle.

[choneofakind]

Give me a lever long enough and a fulcrum on which to place it, and I shall move the world

[red5]

It was my understanding that there would be no math....

[blink]

Originally Posted by Motofool

Why?

If you ride, you didn't give up, you use it each time you take your bike out.

You refill oil and lube your chain to reduce friction, you select your tires and brake pads to increase friction, you balance your CG over the contact patches, you counter-steer, you give gas to increase acceleration and by switching gears, you are applying the old lever principle.

Haha I don't enjoy the calculations part. I enjoy the end result! =]

[BlueHairSar]

Originally Posted by choneofakind

Welcome Andrew.

Good to see that you've met Hernan, our resident physics buff

Jiggles just posted up a video thread designated for nerdy science videos. Feel free to peruse that one as well. It's got some fun stuff in it, like this video:

Link to original page on YouTube.

This is amazing!!!


Lee Parks, David L. Hough, and Keith Code all have good snippets of physics in them, too.

[BlueHairSar]

Originally Posted by BlueHairSar

This is amazing!!!


Lee Parks, David L. Hough, and Keith Code all have good snippets of physics in them, too.

Err, their books do. I'm sure they have good snippets of physics in them, too, though..

[choneofakind]

I got what you meant Sarah.