[amit_purohit20]

Suddenly all the discussions on this thread have ended. It seems either we all are clear about Torque generation and distribution or we have convinced ourselves of the answers given here.

Require more inputs please...

[Jeroen]

Well, maybe some of us just got tired of this discussion

[mallumowgli]

For me official commitments means this thread was moving too fast for me to catch up. I have lots of queries - I think I will be able to put them up only by next FY!!

[Samurai]

Originally Posted by amit_purohit20

Suddenly all the discussions on this thread have ended. It seems either we all are clear about Torque generation and distribution or we have convinced ourselves of the answers given here.

I thought the ball was in your court. But the truth is... the man who led us this path (constant torque drive) is not active for a while. It was an interesting ideal condition to make our overwork our collective brains.

Frankly, this discussion has really clarified my understanding of power, besides the torque. These days when I do my long drives, sometimes I find myself analysing the gear changes I do in the ghats, based on these discussions.

[Sutripta]

Is there anything new to discuss?

Regards
Sutripta

[amit_purohit20]

Originally Posted by Samurai

I thought the ball was in your court.

Oh and I thought I had poured my heart out with whatever little knowledge I have about this topic and was about to infer that you agree with my thoughts!

I have presented my views on the topic and now its upto other members to either agree or disagree. If they disagree with some explanation then again we can have a healthy discussion.

But yes I expect the person to go through the whole thread before triggering the discussion again.

Originally Posted by Sutripta

Is there anything new to discuss?

Regards
Sutripta

Does it mean that you have agreed to whatever I have written in the last 3-4 posts? If not then certainly I would like to know about your view point.

I would certainly say that this thread had made me think hard especially about the analogies and examples which Samurai had quoted. Its so embedded in my brain that now I use this for interviewing the candidates . I think some of the analogies are very good to use to do a quick check on a persons understanding of physics.

I admit that even I have got confused 3-4 times, but that has made me think hard and now have more clarity about the topic.

[Sutripta]

Originally Posted by amit_purohit20

Does it mean that you have agreed to whatever I have written in the last 3-4 posts? If not then certainly I would like to know about your view point.

Go through
http://www.team-bhp.com/forum/4x4-te...ml#post3368249
and decide whether we are in agreement or not!

Regards
Sutripta

[Sutripta]

Originally Posted by Sutripta

Is there anything new to discuss?

We all know how an normal differential works, and the the torque split is equal and therefore 50:50.
Maybe someone can explain (simply and lucidly) how a mechanical (ie not clutch pack/ fluid coupling) differential can split torque non 50:50, while keeping both output shafts at the same speed. Normally used as a centre diff in high performance sedans.

Regards
Sutripta

[amit_purohit20]

Originally Posted by Sutripta

Guys,

Case 1: the wheel is clamped.
Torque transmitted by shaft = torque generated by source.
Velocity =0. Acc = 0.

Let the clamp be released.
If the system has no other force acting on it,
Torque transmitted by shaft = torque generated by source.
The wheel will accelerate using Newtons F= M * A (well, the rotating,rather than linear interpretation).
And will continue to accelerate. To infinite velocity if the force is maintained long enough.

Considering your torque source is a pulley with a woven rope and carrying a weight (constant torque source).

I do not agree that Torque will be the same before and after release of the clamps.

You do not talk about Power here, thats the problem.

If torque on shaft remains same even after releasing the clamps then the equation P= KNT where P- Power, N-rpm and T-torque is violated. Its the shaft rpms that increase which bring the Power figures here.

Also let us go back to strain guaging of the shaft. The setup measures torque on the shaft based on the twist of the shaft.

So before the clamps were released lets say the torque on the shaft was 50 Nm. This was measured as 50 Nm because of the corresponding twist of say 1 degree measured on the shaft.

Here the shaft had got twisted because there was a torque applied from the source and on the other end of the shaft it was resisted by the clamps.

Now when you release the clamps dont you think because of the reduced resistance the shaft will try to untwist itself or I mean that now the shaft will twist less because of the less resistance.

Now as the shaft twists less the strain guages will read less torque than 50 Nm ie that the shaft is expereincing less torque now.

So I ask you a question here, Do you agree that once the clamps are released the shaft will twist less as compared to before the release of clamps?

If you answer yes then we should agree that the torque value drops after the clamps are released.

(Please note that let us not discuss the torque source now. We have already decided that its a pulley with a rope woven on it with a weight attached hanging down. Or in simple terms we say that one has applied 50 Nm on the shaft.

I also do not want to go into acceleration and rpm values now at this stage.)

[DirtyDan]

By definition the motor is a constant torque source in this example. There is always resistance whether it involves clamping the shaft or not since the wheel provides resistance. The only difference I see in clamping or not clamping is that, depending how heavy the wheel is, the shaft and wheel may start to rotate. But the torque to the shaft will remain constant. In this hybrid, idealized situation, whether there is rotation or not, I would expect your strain gages to remain constant, too. Your gages are truly egalitarian. They don't care if resistance comes from clamps or the mass/acceleration of the wheel. No clamp bigots they be.

[amit_purohit20]

We have already inferred that torque is based on resistance.
What I understand from above post is that you are saying that resistance is same before and after clamps are released. After the clamps are released the resistance is kept same by the inertia of the wheel.
Well I dont agree here. Can you explain the same concept using power equation.

Lets take the example of a person standing on the bicycle pedals trying to go up a hill and suddenly the chain snaps/slips. Dont we feel a reduction in resistance? If there is reduction in resistance the torque should fall down.
My opinion is that this torque gets transformed into power thus increasing rpms.
Let me ponder more on your school of thought...

[DirtyDan]

Originally Posted by amit_purohit20

My opinion is that this torque gets transformed into power thus increasing rpms.

Yeah, .... Another way of looking at it is that the torque that was straining against the clamps is, upon release of the clamps, now free to work against the wheel....immediately.

[amit_purohit20]

Originally Posted by DirtyDan

Yeah, .... Another way of looking at it is that the torque that was straining against the clamps is, upon release of the clamps, now free to work against the wheel....immediately.

But this another way of looking means a lot in terms of torque value going up or down.

Ok little bit off-topic but I want to know how the torque and power graph is plotted wrt rpms using a dynamometer.

Consider it a simple prony brake dynamometer, do we force the rotating engine shaft at a certain rpm to stop completely and then note the highest torque value measured?

So to plot the complete graph do we do this at nearly all rpms of the engine and then we have torque versus rpm graph?

Why I ask this is that my opinion is that in our earlier example if we use a strain guage to measure torque we will get a different value but if we use a prony-brake dynamometer we will get a different value of torque.

Reason in the dynamometer case the shaft is again made to come to stand still and thats where the max. torque is measured. Whereas in a strain guage type of setup the torque is measured instantaneously and dynamically wrt to time.

[Sutripta]

Amit, humour me and clear my doubts on a very basic question:

Warning - thought experiment!
Take a cube of metal resting on a frictionless horizontal plane.
That plane also has a post embedded in it.
On two opposite sides two strings are attached.
One of the strings is attached to the post.
The other string is pulled by a constant force away from the post.
(If one wants to be pedantic, the force, line of strings etc is horizontal and passing through the centre of mass of the metal block)
Suddenly the restraining string breaks.

What would be the tension in the other string, both before and after the restraining string broke? And in the restraining string before it broke?

Regards
Sutripta

[amit_purohit20]

Originally Posted by Sutripta

What would be the tension in the other string, both before and after the restraining string broke? And in the restraining string before it broke?

Regards
Sutripta

Lets say the string is being pulled at 10 N. So both the strings have 10 N force applied.

Now if the restraining string snaps the resistance has come down to zero (ideal case) so the tension in the other string cannot be 10 N.

The 10 N force/tension in the string existed because of applied force= 10N and resisted force by the pole=10N and thats why the body was in equilibrium.

Once the reaction provided by the pole was removed due to the snapping of the string the tension in both the strings should cease and be zero (ideal case).

One cannot apply any force to the string if there is no reaction.

Please elaborate on your thoughts.