Tiffys, eggheads or cleverclogs - answers please.

#1
I am a civvy DMI at SEME. I would like to know the in- depth answer to this question.
A Tracked vehicle (let's take the 432) is travelling at 20mph. The Bottom of the track is travelling at 0mph. What speed is the rest of the track travelling at?
I thought it was 40mph at the top for it to 'catch up'. Upon discussing it with another Instructor I now realise that surely the track will still be doing 20mph as would the Final drive and the rest of the running gear.
Please enlighten us with your world of Technical knowledge, expertise and guesswork.
Thanks. Grizz.
 
#2
Is this going to stop you sleeping?

The final drive isn't turning at 20mph its doing a set no of rpm, a point at the edge maybe going at 20mph but a halfway across the radius would only be going 10mph.
 
#3
The track speed is irrelevant. It's the speed of the vehicle moving over the track that is important. The track rotates all at one speed (obviously).

The final drive "pulls" the 432 along, using the track as friction on the road.
 
#4
well seeing how the speedo drive is off the final drive, final drive turns the tracks,

so speed equals the speed the final drive is turning,

track connected to final drive so hey presto track traveling at said speed.

Or is this some kind of elaborate wah?
 
#5
wee_jock_poopong said:
well seeing how the speedo drive is off the final drive, final drive turns the tracks,

so speed equals the speed the final drive is turning,

track connected to final drive so hey presto track traveling at said speed.

Or is this some kind of elaborate wah?
Thats assuming the the speedo works on a ratio of 1:1.
 
#6
If the top of the track was only travelling at 20 mph too, it wouldnt appear to move would it.

It quite obviously has to be going twice the speed of the vehicle over the ground at some point. Unless of course youre on about relative speed. If so, the following simple equation usually works;

v(t)=r'̒ (t)=(x'(t),y'(t),z'(t))=(-a sin(t),a cos(t),b)

:wink:
 
#9
I think you would be better off measuring the track, timing the track for one revolution at 20mph, then using a far simpler formula of speed= distance/time. Then converting from m/s to mph but thats just how i would do it
 
#13
Cant be arrsed.
 
#15
20mph is the speed of the vehicle. Tracks and wheels rotate and are measured in revolutions per second/minute etc. Technically speaking the speed of the track is how fast it is rotating. Assuming track speed is constant for one revolution (which it probably isn’t) all you need to do is measure the circumference of the track and time a revolution (As previously mentioned). Even an A Mech could do the maths to work out answer.

:wink:
 
#16
alb said:
20mph is the speed of the vehicle. Tracks and wheels rotate and are measured in revolutions per second/minute etc. Technically speaking the speed of the track is how fast it is rotating. Assuming track speed is constant for one revolution, all you need to do is measure the circumference of the track and time a revolution.

Geeeeeeeeeeeek!
 
#17
Bearing in mind that it is a "track laying vehicle" the track will be travelling at 20mph to be laid at the rate required for the vehicle to progress along it at 20mph.
 
#18
craftsmanx said:
Bearing in mind that it is a "track laying vehicle" the track will be travelling at 20mph to be laid at the rate required for the vehicle to progress along it at 20mph.

Well done bullet. That wasn't exactly the question though.
 
#19
If you blink fast enough the top of the track actually goes rearwards.....will the relevant speeds be addative, albeit in a subtractive sense, or are they cumulative but with a negative sign...???

If you manage to synchronise your blinking to vehicle speed the whole track then becomes statitionary and therefore impossible to calculate it's true speed, due to the inability of division by zero.
 
#20
UnderTheBreech said:
If you blink fast enough the top of the track actually goes rearwards.....will the relevant speeds be addative, albeit in a subtractive sense, or are they cumulative but with a negative sign...???

If you manage to synchronise your blinking to vehicle speed the whole track then becomes statitionary and therefore impossible to calculate it's true speed, due to the inability of division by zero.

I take it you haven't been CDT'd for a while then, UTB? :wink:
 
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