torque/power distribution on two or more axes

[simone1996]

Good evening to all,
excuse the doubt, probably for many of you.
on the theoretical level, therefore without defining a realistic application, is there a breakdown of the torque/power in case a motor transmits power on two different axes?
from the scheme (perdominates I know that it is not a great deal) the axis b of the engine transmits power to the axis b and c.
What do they see?
the three toothed wheels have the same number of teeth.

 

[Stan9411]

there is no pattern.

Anyway, I think you don’t have a car with integral traction.. the concept is the same

 

[simone1996]

Thank you very much for the answer stan9411!
actually no my car is front traction la
now I try to reshape the scheme

 

[Stan9411]

Well now I see the pattern.

I would say that to explain it with simple words, to govern is always the couple. That puts the two users on you. the balance of forces on a rigid body (in our case the tree/wheel b) always applies, regardless of whether the strengths are one, two or a thousand. so also the couple to the tree b will always be simply the sum (with transmission ratios) of the couples required by the trees to and c.
Now you've reached the engine. always simplifying a lot, the torque dispensed from the engine depends on its rotation regime. the system will go towards its stationary point of regime: in other words the engine will turn the whole system to that rotation regime in which the drive pair equals the torque to the shaft b.
if one can adjust the torque delivery (we will press on the accelerator/stop the current) I will have an excess of torque, in addition to that required by my utilities and the system will be able to accelerate, until you arrive at a new stationary rate.

 

[simone1996]

thank you many, if I have well understood now making an example: a 5nm couple engine
torque seen from u1 + torque seen from u2= torque engine (5nm) or 2.5 nm cada utenza.
In this way, at cascade, the power is equally distributed. Is the reasoning correct?
I suppose I have 1:1 transmission ratio for both axes

 

[TETRASTORE]

torque seen from u1 + torque seen from u2= torque engine (5nm) or 2.5 nm cada utenza.

in your case (Report 1/1) not necessarily the torque must be divided in half, as indicated in a previous post depends on the torque, also differentiated, required by the two u1, u2 users; obviously necessary condition is that the sum of the two should not exceed (if not in particular cases) the torque supplied by the engine.
the same applies to the power linked by direct proportionality to the pair (p(kw)=m(nm) x rpm (rpm) /9550).

 

[Stan9411]

a 5nm couple engine
torque seen from u1 + torque seen from u2= torque engine (5nm) or 2.5 nm cada utenza.

and no.. as it also tells you tetrastore, the couple that dispenses the engine naturally (so without checking it mechanically or electronically) is always the sum of the couples required by the users.. the users “command” the operating condition in an uncontrolled closed loop system

 

[simone1996]

but it is not clear to me. Sorry they're pumpkin.
if I have a motor connected by means of geared wheels with rdt 1/1 to two trees which are identical and I have already defined how much torque to the valley I have to have, and I have already decided that the pair will have to be equal fall tree, then I choose the engine that has as a plate the sum of the pairs present on the two trees?

to say if I block one of the two trees, then the pair if it catches all one tree, but if both are free to be rotated I think the pair is broken in a fair way.
Is my reasoning correct or am I still touching?? Thank you.

 

[Stan9411]

if the two utilities require the same pair, the one coming from the engine will be distributed exactly in half.

if u1 requires 10 nm and u2 it requires 2, the engine will erode 12, but of course 10 will go to the shaft of u1 and 2 to the shaft of u2.
in that sense: “the utilities command how the couple is divided”.

I didn't understand why you didn't understand....

 

[Stan9411]

to say if I block one of the two trees, then the couple gets it all one tree

If you block one of the shafts and have a rigid transmission (type gears) it blocks the whole system.. If you have for example a trapezoidal belt, that will slide on the blocked shaft and the motor will only erode the couple to accelerate the free shaft until the speed in which strong couple and drive pair of equals.
the motor, unless you control it in pairs, always brings in balance with the load.. If you force it to give more torque than the load requires, it means that the whole system has to accelerate.

 

[Stan9411]

crashes the whole system

of course all the system even if it does not turn remains “in couple”. . so the teeth of your pair of geared dense wheels will go under stress and in the worst case they will break .. or if you have a strong beautiful transmission, it will be the motor to go under stress and will turn off ... as when you take a rough climb in fifth to 70 per hour ... ask the torque engine that does not have, brontola, borbotta and if you do not scale a couple of marches, it turns out.

 

[simone1996]

ah ok now I have well understood, you are right the system is rigid! It can't work as I said.
thank you many have cleared my ideas and learned something new.