27 replies to this topic

[MatsNorway]

I have calculated the acceleration of a rolling wheelset to be 0.06506 m/s^2

Now im wondering about how i add the rolling friction to get the new acceleration..

Speed after two meters on the slope is then 0.509 m/s

Rolling friction is set to be: 0,0005

Wheelset weight is 1500kg.

Edited by MatsNorway, 22 February 2012 - 14:26.

[Wolf]

MatsNorway, on Feb 22 2012, 15:20, said:

I have calculated the acceleration of a rolling wheelset to be 0.06506 m/s^2

Now im wondering about how i add the rolling friction to get the new acceleration..

Speed after two meters on the slope is then 0.509 m/s

Rolling friction is set to be: 0,0005

Wheelset weight is 1500kg.

If I am not mistaken, acceleration due to friction will be f.rf * gravity which is roughly 0.005 m/s^2. Total acceleration would then be 0.060155 (substracting rolling from calculated), and speed after 2m 0.491.

EDIT: I overlooked the mention of the slope, which reduces normal force- then the acceleration from rolling friction will be f.rf * gravity * cos (alpha), where alpha is angle of the slope...

Edited by Wolf, 22 February 2012 - 16:46.

[MatsNorway]

What is f.rf?

Force rolling friction or something?

or are you meaning to say force * rolling coefisient * gravity * alfa etc.

[Wolf]

I meant coefficient of rolling friction- acceleration is force/mass, and force is rolling resistance coeeficient*mass*gravity*cos(alpha), so the resulting acceleration is coefficient*gravity*cos(alpha).

[Lee Nicolle]

Now being practical you can have 2 wheels the same weight, but a heavy outer [tyre] will accerate slower than a lighter outer.It is called centriigual mass. A larger diameter will accelate slower for the same weight too ofcourse.

[MatsNorway]

Not a issue with train wheelsets.

[bigleagueslider]

MatsNorway, on Feb 22 2012, 06:20, said:

I have calculated the acceleration of a rolling wheelset to be 0.06506 m/s^2

Now im wondering about how i add the rolling friction to get the new acceleration..

Speed after two meters on the slope is then 0.509 m/s

Rolling friction is set to be: 0,0005

Wheelset weight is 1500kg.

MatsNorway,

Wheels or tires have rolling resistance, which is a bit different than simple sliding friction. Rolling resistance is mostly due to hysteresis losses.

The power to overcome rolling resistance at a given forward velocity is approximated by C x N x V, where C= coeff. of rolling resistance, N= wheel normal force, V= forward velocity.

[Wolf]

I thought it can be approximated by a constant at the speeds involved- the final speed in this example is 1.8 kph (and Mats has given it as a constant, so my suggestion was based on that)...

[MatsNorway]

you guys are confusing me...

the slope is 0.5 degrees.

The wheel stands in it.

the wheel gets dropped.

and it rolls down 1.5meters. it then comes down the "hill" to a flatt railroad track that goes 5 meters.

whats the speed at the end of those 5 meters?

the formulas is actually the most important thing here. as i might need to alter the "hill"

Im guessing its easiest to splitt the calculations. one for the hill one for the straight.

Edited by MatsNorway, 28 February 2012 - 07:07.

[Wolf]

HTH

[MatsNorway]

Thats very very very nice. Thank you.

[REN_AF1]

MatsNorway, on Feb 28 2012, 15:46, said:

Thats very very very nice. Thank you.

Real off topic, but inspired by the Mathcad sheet above: Have you ever tried using SMath.. Brilliant application, will even run without install and on mobile devices to.

http://en.smath.info/forum/


R (not affiliated at all)

[MatsNorway]

I have it installed but have not tested it yet.

This is the mechanicm that will stop the wheels from rolling. I am currently unsure about how strong the spring needs to be.

The track prior is 5-6meters with 0.5 deg slope.

The idea is to have a spring so that if the power drops out or a purely air pressured system fails.. this will still stay up and stop 5-6 wheels from going into the gate and into the workshop.

An alternative is to have a locking mechanism that needs to be removed to lower the stopper arms. i could ofc have both.

http://i1188.photobu...pringreturn.jpg

Initially i thought i needed a spring about the strength of the weight of the wheelset but i think thats a bit overkill.

6 wheels combined gives me a planar force about 770N so the spring in the picture is 1481N. but that value is static. one lone wheel picking up speed could surpass(?) that load.

Right now im thinking that im going to combine the lconnector joint for the spring and pneumatics with the stopper arms. and move the SKF housings further inn. saves me one keyway in center and that part around it.

Edited by MatsNorway, 07 March 2012 - 09:28.

[MatsNorway]

second gen as promised previously


Then i came up with this one.





BUT!

Now they say that the wheelsets can`t hit each other. the shock from the wheelset hitting each others will go into the bearing....

hufff.... i actually got really dissapointed for this. Because this idea is a good one. And it looks good too.

I "stole" the prinsiple from a BIP turntable. added stuff and woila it was complete. took me basically 7-8 hours to design it.

its pneumatic. And uses gravity as a part of the failsafe. it also needs normally open valves to release pressure under a power loss.

outer pieces is designed to be dismountable, there is a "slot" where the feet welded on the beams goes into. This takes the load. the beams are casted into the ground.

The thingy unde the cylinder is a damper, also used on BIP turntable. i might have had to consider a different one due to bigger loads and weather.

Edited by MatsNorway, 30 April 2012 - 13:54.

[gruntguru]

Lee Nicolle, on Feb 23 2012, 18:09, said:

Now being practical you can have 2 wheels the same weight, but a heavy outer [tyre] will accerate slower than a lighter outer.It is called centriigual mass. A larger diameter will accelate slower for the same weight too ofcourse.

Mats, you can't ignore the above post.

A significant portion of the energy goes into rotatiion, eg a pipe rolling down a slope does not accelerate as fast as a frictionless brick sliding down the same slope.

[MatsNorway]

Yes i can and its irrelevant anyway now.. there will be 98% cargo wheelsets and they got no gear and such.. And when a train wheel banks a corner the wheelprofile makes the effective radius on the outer wheeldisk bigger as well. Unless they are very worn.

as said this is all irrelevant.. i have to figure out a new machine that does the transport without crashing the wheels into one another.

[Tony Matthews]

MatsNorway, on May 1 2012, 12:27, said:

...when a train wheel banks a corner the wheelprofile makes the effective radius on the outer wheeldisk bigger as well.

Is that what the taper on the 'tread' does? I like your renderings, they're cool.

[MatsNorway]

Tony Matthews, on May 1 2012, 13:19, said:

Is that what the taper on the 'tread' does? I like your renderings, they're cool.

Yes thats why there is no need for diffs in trains. But the flange is just a safety device. its not supposed to touch. renderings are the fun stuff we usually never get to do..

But the profile alone is not the only device that aids in turning. lots of fancy bushings and links to make things go as they desire.

[gruntguru]

MatsNorway, on May 1 2012, 21:27, said:

Yes i can and its irrelevant anyway now.. there will be 98% cargo wheelsets and they got no gear and such . .

You can't ignore moment of inertia when calculating acceleration. The final speed of a bare wheelset at the bottom of a hill will be a lot slower than a heavy wagon using the same wheelsets at the bottom of the same hill.

[MatsNorway]

Its ok.. no wagons here.

And the total energy will be the same will it not?

And i went talking to my prosject leader. the wheeldisc to wheeldisc contact is allowed. its only a matter of stopping the "stacking" of wheelsets to prevent wheeldiscs from hitting bearings and such.

So this thing is still on.

Edited by MatsNorway, 02 May 2012 - 06:11.

[gruntguru]

Yes, same total energy. Just saying the speed of a wheelset at tyhe bottom of the slope will be significantly less than calculated if you neglect rotational KE.

What does "stacking" look like?

[MatsNorway]

gruntguru, on May 3 2012, 00:06, said:

Yes, same total energy. Just saying the speed of a wheelset at tyhe bottom of the slope will be significantly less than calculated if you neglect rotational KE.

What does "stacking" look like?

Will the impact forces be the same?

Stacking was perhaps a bad wording..

This is risky for bearing life. (best i could find)



This is good and correct storing of train wheels.

Imagine one truck driver crashing into this pile if they where stored like above. all bearings gets destroyed. one bearing is what??? 1000-3000$?

Edited by MatsNorway, 03 May 2012 - 14:57.

[gruntguru]

MatsNorway, on May 4 2012, 00:56, said:

Will the impact forces be the same?

Probably lower - some of the energy stored in rotation might be dissipated during impact through sliding friction on the rails. Interesting question.

[Greg Locock]

The linear momentum will be lower so the impulse will be smaller.

[gruntguru]

Interesting. Reminds me of the (near elastic) collosion of two billiard balls - one stationary and one moving.

1. If the moving ball is rolling, it will follow the other ball after the collision. ( < 100% transfer of momentum.)
2. If the moving ball is not rolling (eg a 2-ball Newton's cradle) it will stop dead after the collision. ( = 100% transfer of momentum.)
3. If the moving ball has sufficient backspin, it will bounce backwards after the collision. ( > 100% transfer of momentum.)

So the collision between wheelsets should be type 1?

Edited by gruntguru, 05 May 2012 - 04:43.

[Greg Locock]

I doubt there's much energy lost in the collision, as a %age of the total, so you'd be very close to billiard ball physics.

Unfortunately it is very difficult to work out what you are really interested in, the peak force, unless you know a lot more about the system than just masses and velocities. I'd make a wag that the effective stiffness of the collision is of the order of 50 kN/mm and take it from there.

[bigleagueslider]

I don't know if the polar inertia of the moving wheelset would have much significance, since the wheelsets have equal diameters, similar mass center offset, and would contact at a point essentially in-line with both body's mass centers. The only transfer of rotational momentum would be that due to friction when the wheel ODs briefly touch at the initial contact. It would mostly be a transfer of linear momentum, and this would be a fairly efficient process due to the high structural stiffnesses of the wheelsets.

[MatsNorway]

I discovered late, but in time that if you pull a line between the center of the wheelset and the arms contact point it should pass over the center joint of the arm. That way it will be pushed backwards rather than down. the lump below is a damper to take the forces of up to 3.2 tonnes (2 wheelset)s coming in at max speed of 1.23m/s

The tiny rod just above the damper is a spring who is there to push the wheelset back (after being stopped by the damper) in order to maintain damper functionality when more wheelsets stacks up behind.

Correct angle

Incorrect angle


Whe have not been doing any math on the forces that goes into the center joint because we are unsurtain about the impact forces. Mostly due to the damper. Linear energy damping? peak at the start due to the high speed? most likely. But how much..

4tonnes static in FEA gave good results.

Edited by MatsNorway, 11 June 2012 - 06:52.