cvt belt question

[wangp1283]

One of the biggest problem with the CVT belt is that it cannot handle a
great deal of torque.

This problem seems pretty simple to solve since all it takes is to spin
the belt at a higher rpm for a given amount of power. For any power,
just spin the belt at whatever high rpm it takes to decrease the
torque, and then bring the rpm down. Just use gear ratios. So why not?

can someone
explain to me for once and for all?

by the way, what's the typical efficiency for the belt?
Thank You.

 

[patprimmer]

I expect there might be a lot more inertia in the drive train, and extra friction losses in the step up and step down gears, thereby decreasing performance to a greater degree than the gains with a CVT.

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[wangp1283]

Inertia might be a factor. Any other reasons? such as centrifugal forces? I heard the belt may unload if the centrifugal forces is too high.

But anyway, check this out. (going off on a tangent)

http://old.lib.ucdavis.edu/pse/cvt04/papers/04CVT-03.pdf

From this, it turns out the "shifting force" and the "clamping force" for a belt/chain CVT are NOT the same thing.

Is the "shifting force" the force it takes to actually move the pulley apart/closer?

And I would assume the "clamping force" is the force it takes to create the friction needed.

I've always thought they are the same thing. How can the hydraulic actuators create 2 different forces at the same time?

And if they are indeed two separate forces, which one contributes to more loss? the shifting mechanism or the clamping mechanims?

Can someone explain please?

 

[wangp1283]

I think I've found the answer to this. It's mainly due to the centrifugal force. A simple physics calculation shows that for an endless chain moving at constant velocity with mass density/unit length of P, and velocity V, the Tension T = PV^2.

This mean even if you step it up with a factor of 2, the centrifugal force would quicky amount to thousands and tens and thousands of pound, due to the exponential factor.

Therefore, you will soon defeat the purpose of "stepping it up", which is the decrease the tension in the first place.

 

[adfka]

it takes energy to bend and unbend the belt. which causes heat. so the faster it goes around, the more it bends, and the less efficient it is.

 

[808state]

I belive the actual "Load" of the equation is not being considered here. the amount of friction that a CVT belt is subjected to when "flowing" up to a higher gear ratio is much more considerable than downshifting. Although, friction does uccur when downshifting, but seems irrevelnent since it aids the drivetrain to slow down.
When the hydraulic preassure is increased to the pulley halves to make a change in ratio, more preassure is added from both sides of the belt. Friction from the pulley and belt has to be 100% to avoid a damaging occurance of "belt-slip".
Ive gathered information on CVT's for several months and for high torque/rpms applications, belt drives are slowly catching up to cone, disk, other CVT systems.

Hoorah Grease Monkey

 

[tbuelna]

Limits of RPMs with push belts?

It's the mass of the belt. You must realize that the "belt" is actually a heavy chunk of steel plates and bands. The centrifugal forces it encounters, which tend to unload it, increase at the square of its pulley speed. Thus, there are practical limits to pitch line velocities achievable with a push belt variator. Remember, HP capacity increases linearly with speed, torque capacity increases linearly with pitch line radius, but centrifugal forces (and hoop stress) increase exponentially with RPM's.