|
Post by q8 on Jan 12, 2006 18:09:07 GMT
Someone has said to me that if a train is going along at 40mph the wheels are going much faster. I am not sure I follow the logic of that.
|
|
|
Post by Chris W on Jan 12, 2006 19:08:38 GMT
Bob I think I can see where the person who said this is coming from. As the wheels turn they appear to be traveling faster than the main body of the unit... this is because they are obviously far smaller than the body and as a train moves the tread of the wheel moves from the track towards the body and then forward and back down towards to track to complete the cycle!! I think the person who told you this has either got confused with the illusion of the wheels turning faster than the train or has simply been taken in, but to me..... NO - the wheels travel at exactly the same speed as the train ;D ;D Best wishes Chris
|
|
|
Post by q8 on Jan 12, 2006 20:23:02 GMT
No, you misunderstand Chris. This geezer insists that in terms of wheel revolution the tyre is turning at several times the actual train speed. I can't follow that reasoning at all. If the train is going 40 mph the rail is passing likewise at 40mph so how can the wheel tread be going faster?
|
|
Chris M
Global Moderator
Forum Quizmaster
Always happy to receive quiz ideas and pictures by email or PM
Posts: 19,397
Member is Online
|
Post by Chris M on Jan 12, 2006 21:29:52 GMT
Don't wheels on teh outside of a curve travel a bit faster than those on the inside or something
Phil, I think this is one for you....
|
|
Deleted
Deleted Member
Posts: 0
|
Post by Deleted on Jan 12, 2006 21:31:15 GMT
If you think about it, the bottom of the wheel tread is stationary: it is in contact with the rail and therefore it cannot be moving (unless there is wheel slip).
It therefore follows that the top of the wheel tread must be moving at twice the speed of the train.
|
|
Deleted
Deleted Member
Posts: 0
|
Post by Deleted on Jan 12, 2006 21:39:47 GMT
Don't wheels on teh outside of a curve travel a bit faster than those on the inside or something Phil, I think this is one for you.... No - as the axels on pretty much all trains are fixed. If they had a Diff' on the other hand, the outer wheel would travel faster. Sam
|
|
Tom
Administrator
Signalfel?
Posts: 4,099
|
Post by Tom on Jan 12, 2006 21:42:11 GMT
Don't wheels on teh outside of a curve travel a bit faster than those on the inside or something They certainly have further to travel (hence when measuring track you should use the negative rail as it's roughly in the middle), however they can't be going any faster as they're still connected to the wheel on the inside by the axle. Or have I lost the plot?
|
|
|
Post by Admin Team on Jan 12, 2006 21:44:21 GMT
Oh no - we're getting into physics here I think!
I too seem to have heard stuff about this; not necessarily aboout trains, but the pros and cons for low profile tyres on cars etc IIRC.......
stops before making an idiot of himself
|
|
|
Post by Chris W on Jan 12, 2006 22:20:37 GMT
If the train is going 40 mph the rail is passing likewise at 40mph so how can the wheel tread be going faster? in a very deep voice"Gentlemen - we have entered: THE TWILIGHT ZONE do doo do dooooo..... do doo do dooooo" ;D ;D ;D WHAT THE HELL IS GOING ON...!!!! Bob - as you say if the train is going 40 mph in relation to the rail, the rail is also travelling at 40mph in relation to the train, but it doesn't make a combined 'impact' speed of 80mph as the rail is stationary... unless of course its another 92-stock taking half the surrounding equipment with it ;D ;D or away from it!!! The wheels turn at the the same speed as the train is travelling surely!!! You've now overloaded my brain - ok not difficult - but if you guys carry on this that I'll have to concede to my g'friends' call to go to bed !!! And then probably be unable to sleep as I'll be thinking about this ;D
|
|
Deleted
Deleted Member
Posts: 0
|
Post by Deleted on Jan 12, 2006 23:40:28 GMT
i can see what sydneynick is talking about - just need to think about it a bit before you can get your head round it, but his explination makes sense to me.
|
|
Phil
In memoriam
RIP 23-Oct-2018
Posts: 9,473
|
Post by Phil on Jan 13, 2006 0:07:28 GMT
Ok: we'll try to do this in one.
The wheel has TWO speeds: rotating speed and speed through the air.
So, at any point the centre of the axle is going at the same speed as the train (through the air). At the same instant the bottom of the wheel is touching a stationary rail so its speed through the air is zero. Hence by simple arithmetic the point at the top of the wheel must be going at twice the speed of the axle (through the air) and hence twice the train speed.
All this time the rotating speed of the wheel is constant, and depends solely on wheel diameter, but (and this is where it DOES become confusing) if you follow a fixed point on the wheel (as if it were painted with a blob) its speed through the air (if drawn on a graph) would look like a series of viaduct arches (too late at night to draw the graph, scan and upload it, but I'll do it if wanted tomorrow) The flange of the wheel (the bit below the rail ) is actually going (slightly) in reverse.
And ignore all the stuff about outer/inner wheels and differential action: that's something else entirely and we did it in another thread a couple of weeks ago
|
|
|
Post by q8 on Jan 13, 2006 6:31:56 GMT
Now I am TOTALLY conflummoxed. [goes back to bed as my brain hurts]
|
|
|
Post by compsci on Jan 13, 2006 7:40:40 GMT
The train, and hance the wheel axle, is moving at a certain speed.
Relative to the wheel axle, the edges of the wheel are moving at the same speed.
If it is moving to the right across your screen, the wheel will be rotating clockwise, so the top goes right and the bottom goes left.
From the point of view of a stationary observer, the bottom of the wheel is moving at speed right + speed left = stationary.
The top of the wheel is moving at speed right + speed right = double speed.
|
|
|
Post by stanmorek on Jan 13, 2006 9:49:17 GMT
Rotational speed is different to to compare the two is meaningless.
The outside tread of the wheel is travelling at the same speed as a fixed point on the train, i.e. it covers the same distance over same length of time (assuming no slipping). However, a fixed point close to the centre of the wheel is travelling slower.
I think people are getting confused between linear and circular motion. Imagine a circle spinning at a constant angular velocity, i.e. it is spinning in such a way that it completes a whole turn over the same length of time every time. And there are two points on the circle, A and B. A is near the edge and B is near the centre. After a complete turn of the circle, A has travelled further than B in a linear direction. [Circumference=2 x pi x radius]. Because this happened over the same length of time, A is travelling faster than B but the circle remains at the same rate of spin. A speedometer calculates speed of a car by effectively multiplying the circumference of a wheel by the angular velocity.
Take the example of the train going round a curve, as Russ kindly reminded me that the wheels have a coned profile (a tread of differing diameters). The axle rotates at the same angular velocity but inner wheel travels on the smaller diameter tread, hence a shorter distance, and the out wheel travels on the larger diameter tread, hence the longer distance.
|
|
Deleted
Deleted Member
Posts: 0
|
Post by Deleted on Jan 13, 2006 9:56:20 GMT
Don't young people learn applied maths anymore? Wot, squire??
|
|
Phil
In memoriam
RIP 23-Oct-2018
Posts: 9,473
|
Post by Phil on Jan 13, 2006 11:38:38 GMT
Don't young people learn applied maths anymore? Sadly not StanmoreK : since you left school, and the introduction of the all-singing, all-dancing AS levels, circular motion (other than the v=rw bit) has been considered too difficult for the poor darlings. This is all down to the dilution of maths GCSE, and so on. Derivation of a=v 2/r etc has all gone. Hence the level of my explanation above: what you said is technically right but would confuse almost all the readers - especially people like the t/ops who never did A level maths The whole problem in this thread is that the original post was made on a vague statement from a third party. The answer (as StanmoreK says) DOES depend on whether the original question was regarding linear or circular motion of the wheels.
|
|
|
Post by edb on Jan 13, 2006 12:36:28 GMT
I did mechanics during my engineering foundation year and my first and only year of mech eng.
Never got all of it. Thought most of what you says ring a bell. Will dig out Applied Mechanics and take a look
|
|