| **another pop quiz (cycling related - sorry)** | DougSloan
*Jul 10, 2002 9:19 AM* | | I've heard this and intuitively it makes sense, but for the life of me I can't figure out how to prove it.
A bike wheel rolling on pavement without slipping is going 20 mph (at the axle). With respect to the road, the bottom of the tire has no velocity, and the top is going twice the speed of the axle (bike). (Also, is this true for the intermediate points betweent the outer surfaces and the hub, or is the speed proportionately less or more?)
Can someone explain this in understandable mathematical terms? Thanks.
Doug |
| **Pop quiz? No. That's an essay question!.................nm** | 128
*Jul 10, 2002 9:24 AM* | | |
| **yes, and show all work :-) nm** | DougSloan
*Jul 10, 2002 9:42 AM* | | |
| **re: another pop quiz (cycling related - sorry)** | mr_spin
*Jul 10, 2002 10:37 AM* | | As you move outward along the radius, the circumference of the circle gets exponentially larger. Remember PI R SQUARED? (In simple terms, the "squared" part is an exponent, which makes this by definition exponential).
Since a greater distance (circumference) is covered in the same amount of time, you can say that speed increases exponentially the further away from the axle you travel. |
| **does this answer the question?** | DougSloan
*Jul 10, 2002 10:39 AM* | | Am I missing something? I understand the concept you mention, but how does this apply? |
| **what was the question?** | mr_spin
*Jul 10, 2002 11:14 AM* | | What you posted doesn't completely make sense to me, so I did the best I could and explained the mathematics you need.
In the absence of friction, it is not possible for the bottom of a perfectly balanced wheel to have a different velocity from the top. So I don't understand what you wrote. |
| **sorry** | DougSloan
*Jul 10, 2002 11:44 AM* | | I want someone to prove that the top of the wheel is moving at twice the speed of the bike with respect to the ground. That's the guts of it.
Doug |
| **no proof, but...** | mr_spin
*Jul 10, 2002 12:06 PM* | | Check out this site. It's got all kinds of interesting info: http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/bicycle.html |
| **circumference increases linearly--2*pi*R.** | Leisure
*Jul 11, 2002 2:59 AM* | | |
| **mmmmm...donuts! (nm)** | JS Haiku Shop
*Jul 10, 2002 11:22 AM* | | |
| **long answer with formulas** | nn23
*Jul 10, 2002 2:28 PM* | | wrt to the hub
~~~~~~~~~~~~~~~
All points on the wheel have a constant magnitude of velocity around the hub.
|v| = r * w
|v| -> magnitude of velocity
r = distance from the hub
w -> omega, angular velocity which is the rotation speed
Speed can be broken into a horizontal and vertical speed
Horizontal speed Vx = r * w * sin(A)
Vertical speed Vy = r * w * cos(A)
'A' is the angle of spoke wrt to horizontal in direction of movement of the cycle.
Since the wheel is moving wrt to the ground at 20mph, speed of any point wrt ground is
Vx = 20 + r * w * sin(A)
Vy = r * w * cos(A)
Absolute magnitude of velocity
V = under root of (Vx squared + Vy squared)
Putting values in the formulas above we get
When r = radius of wheel r * w = 20 mph (the wheel is not slipping)
Thus at the 12'o'clock position,
Vx = 20 + 20 * sin(90) = 20 + 20 * 1 = 40
Vy = 20 * cos(90) = 0
V = 40 mph
6 'o' clock position
Vx = 20 + 20 * sin(-90) = 20 + 20 * -1 = 0
Vy = 20 * cos(-90) = 0
V = 0 mph
3 'o' clock position
Vx = 20 + 20 * sin(0) = 20 + 0
Vy = 20 * cos(0) = 20
V = 28.3 mph (appox)
For a point on the spoke halfway between the hub and rim, and at 30 degree angle, the speed would be
Vx = 20 + (20/2) * sin(30) = 20 + 10 * 0.5 = 25
Vy = (20/2) * cos(30) = 10 * .866 = 8.66
V = 26.5 mph (appox) |
| **wow, thanks - now my philosophical answer** | DougSloan
*Jul 10, 2002 2:56 PM* | | That looks good, except the velocity at the 3 o'clock position. Is that horizontal? If so, why isn't it 20 mph (the same as the bike)?
Here was my "philosopher's" reasoning: The horizontal speed at the axle is 20 mph. The horizontal speed of the bottom of the tire wrt the ground is zero. Since the entire wheel must remain together (in tact), moving over the ground, the speed of the axle must be somewhere between the speed of the tire at the ground and the top of the tire (at 12 o'clock); since we know the axle is in the middle of the wheel, it must be half way between zero and 40 mph, if the axle is going 20 mph. Right?
Doug |
| **wow, thanks - now my philosophical answer** | nn23
*Jul 11, 2002 10:32 AM* | | > That looks good, except the velocity at the 3 o'clock
> position. Is that horizontal? If so, why isn't it 20mph
> (the same as the bike)?
The point at 3 0 clock position is horizontal. This point is not only going forward at 20mph but also going down towards the ground at 20 mph. Thus actual speed is higher. |
| **good one! nm** | Leisure
*Jul 11, 2002 3:02 AM* | | |
| **"Wheel go round in circles,...** | OutWest
*Jul 10, 2002 9:39 PM* | | ...Billy fly high like a bird up in the sky" Man, that is an old song. I can't even remember who sang it now, wasn't Billy Preston was it? What year? Oh wow, I'll have to wait for the next flashback to find out! |
| **Yup, pretty sure it was Billy Preston, can't remember when. nm** | OutWest
*Jul 12, 2002 6:37 PM* | | |
| **Perfect Circles vs Wheels - quick answer** | jose_Tex_mex
*Jul 11, 2002 7:34 AM* | | I'll have to check out the top part. But what you said about the bottom part having zero veolicty probably goes back to a discussion of static vs dynamic friction. In the case of a perfect circle rolling along the ground without slipping we use static friction. This irks most people as the wheel is moving - why not kinetic? The answer is that the perfect circle only comes in to contact with the ground at a single point (in theory). Since there's no continuum of points involved we do not talk about motion. Wheels have contact areas so I doubt we can bring this arguement in to real life.
Other question would be are you talking about angular velocity or overall straightline velocity? Furthermore, are you looking at the 2 dimensional profile of the bicycle or from the top and considering it in 3D? Finally, were you really interested in the velocity and not just speed. If so it might just be a vector thing.
Here's some more counter-intuition
The rear wheel travels less of a path than your front (remember ridnig through puddles) yet is able to keep up with the front during equal time intervals. Since we have time as a constant we can say that velocity is proportional to distance. Increase one and you have to do that to the other. Since the rear has a decreased distance it follows that it needs a decrease in velocity. Thus, if you look at the overall motion of the bicycle you rear wheel actually has to travel slower. Hmmmm. |
| **How fast do your feet move when walking?** | jose_Tex_mex
*Jul 11, 2002 1:21 PM* | | Although, I am sorry to say I no longer have the article I read sooo long ago, I do remember the results.
A person who walks at say 4mph will actually have their foot move (at the fastest part of their stride) at about 30mph!!! Seems pretty fast huh?
I'll have to find the proof of this. Anyone out there a foot doctor? |
| **How fast do your feet move when walking?** | DougSloan
*Jul 11, 2002 1:50 PM* | | but their feet move at an average of 4 mph, right?
how long do the feet spend "doing nothing" between the 30 mph spurts?
Doug |
| **philosophical answer** | nn23
*Jul 11, 2002 4:13 PM* | | If each foot is moving for only half the time the average speed of each foot for the time it is moving is 8 mph.
However when we are walking, about 20% of the time both feet are planted on ground. Thus time for which each foot moves reduces by 20%. Thus average speed while moving is
8 / 80% = 10 mph.
If the foot were to move like a pendulum and gain speed for half a step and lose speed half a step (OK now this is over simplification) then average speed for each half would be the same and be half of peak speed. Thus peak speed would be 20 mph.
Well atleast we are in the ball park.
Now here comes a twist:
If the person were sprinting at 4 mph then he would have both feet off the ground for about 30% of the time and then max leg speed would really be only (8 / 130 %) * 2 = 12mph.
How does one "sprint" at 4 mph ? |
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