# The moment a clutch begins to engage...

#

**31**Tech Regular

But indeed, this was "3 Havo" material...

I will watch out for your revenge....are you offended now?

Really like that Blis is thinking about the clutch in this way, it`s part of the hobby for some.

If you learn this kind of theories, you understand how difficult the life of a trained engineer is at the local club with all the theory hero`s...lol

But it`s all good in the end...

#

**32**But indeed, this was "3 Havo" material...

I will watch out for your revenge....are you offended now?

Really like that Blis is thinking about the clutch in this way, it`s part of the hobby for some.

If you learn this kind of theories, you understand how difficult the life of a trained engineer is at the local club with all the theory hero`s...lol

But it`s all good in the end...

The objective of this thread is to gather as much information and mis-information to share with everyone. Ive been chatting on RCtech for years now and clutches are spoken of so infrequently. When I first went to the race track, they said ... "it's all in the clutch!" So I spent many years observing, asking questions and hoping to find the answers. The more I learned the more I realised how much I didnt know.

So to all that have information they wish to share, whether correct or not, please submit it so we can all brainstorm and remove the myths and use the collaboration of many minds. After all, the clutch drives the transmission, not the car and it's not like it should be kept a secret because there are so many other engineering aspects to a IC car that have to be mastered before you can get around the track like Alexander Hagberg.

So to clarify and perhaps make it simpler by not hinging the throws on the flywheels pins and make them float. * Because I definitely cant get my head around the hinged forces of a throw

If the throws were fully floating, would the throws have more force to push against the clutch plate if the flywheel is smaller in diameter or larger? I cant get my head around how the tangental forces work but the formula I posted seem to suggest the smaller the radius, the greater the force?

Im thinking of sitting on swiveling chair with my arms extended or spinning a whistle strapped on my finger, the force on my finger seems greater as the radius gets smaller. And newtons said to any force there is an equal and opposite force.

The real challenge is perceiving these forces in three dimensions.

1) The Centrifugal / Centripetal (Newton said we cant have one without the other)

2) The pivot on the hinge ( The deformation of the force applied in two directions ) Hinged on the pin and ramp on the flywheel.

3) The Ramp of the flywheel ( The friction, mass, angle )

So as I said, in the preface of the first post, it's like trying to perceive molecular science and I know that any form of training over many years makes it easier. So any engineers that want to contribute to the facts will have my respect and I would appreciate it being shared.

#

**33**Now the track/forum hero`s stay quiet, right Martin..?.lol..

You guys are on the right track, the formula can be used, it`s simple mechanics/dynamics.

In the formula for centripetal acceleration, we need to use the opposite reaction acceleration called the centrifugal acceleration. Multiply with the mass and we have a centrifugal force, seen in the picture in red arrow.

The speed in the formula is the tangential speed shown with v in the picture,

If we place the picture on the clutch we can understand the parameters

So, a higher centrifugal force.

This force is converted by the angle in the flywheel to a axial clutch force, in the clutch the radius and speed are related, a bigger gap means a hogher axial force, means a shorter slip time.

The spring works against the axial force and is used to control slip time.

Thats it..

You guys are on the right track, the formula can be used, it`s simple mechanics/dynamics.

In the formula for centripetal acceleration, we need to use the opposite reaction acceleration called the centrifugal acceleration. Multiply with the mass and we have a centrifugal force, seen in the picture in red arrow.

The speed in the formula is the tangential speed shown with v in the picture,

**so if we have a bigger radius of rotation of a infinte small piece of mass, we automaticly have a higher tangential speed of this piece of mass.**If we place the picture on the clutch we can understand the parameters

So, a higher centrifugal force.

This force is converted by the angle in the flywheel to a axial clutch force, in the clutch the radius and speed are related, a bigger gap means a hogher axial force, means a shorter slip time.

The spring works against the axial force and is used to control slip time.

Thats it..

But here is where I'm struggling to understand... Because...

Hypothetical :

Lets say we are holding RPM constant at 11,000 rpm, if the throw is half way up the ramp and the force against the plate and spring equalises and the throw doesnt move further up the ramp. The V^2/R does not seem to be a linear formula and it has been said that spring forces are linear.

With the throw half way up the ramp, the radius is greater. The throw itself is now moving more distance, as the radius has increased that means V has increased!!

(I so much prefer the digital world where the simplicity of base 2 maths applies!)

#

**34**
#

**35**Tech Regular

iTrader: (13)

Ok, Dusting, so I assume that if you are using a high bite shoe and there's very little slip it may do similar to marbling soft tyres. Would engagement be better just prior to the power band? So it's a good example of the questions as to why and where the variables play. Aluminium bells, versus steel bells and inertia could be factored in too. I often look at the holes in the shoe to see if they have stretched and if Im stressing it.

I'm enjoying this tread but reading it makes me aware of my limited knowledge of most things RC.....so much to learn

#

**36**Tech Regular

Thats the funny part, it`s been said before, to go fast we dont need the knowledge, just build the clutch in precisse way and learn from experience..!!

Just dont talk about the the theory, because 99% of the trackstories are incomplete or wrong...lol.

But ok, what wins on sunday sells on monday..

#

**37**hold a brick in your hand and swing it in a circular motion at moderate speed.

notice the weight in your hand.

Then, swing it as hard as you can : what happens with this weight??

It will get larger(more?), but the brick is the same (weight) and the radius doesn`t change.

so at the same radius rotational speed will increase weight from a certain mass.

This mass will aply a force forward because it is moving at a slope (the flywheel).

This forward movement is also a mass(clutchflyweights) wich aplies force but now forward against the clutchbell.

Increase the radius and the mass will have more rotational speed (and gain weightforce) with the same axis speed.

So its theory and who cares.

PS all clutches work, just tune them right

#

**38**Tech Regular

So acceleratin is higher, remember s=0,5 a t^2

Distance a stays the same so if a is higher t must be shorter.

If distance is the same and t is shorter, v must be higher...!!

But who cares, race it...!!

If the car is working for lets say 80% optimal, the driver is far more important..

champions benefit from a perfect car.

#

**39**So acceleratin is higher, remember s=0,5 a t^2

Distance a stays the same so if a is higher t must be shorter.

If distance is the same and t is shorter, v must be higher...!!

But who cares, race it...!!

If the car is working for lets say 80% optimal, the driver is far more important..

champions benefit from a perfect car.

See you at the track ( now you are planning to race the big boys)

#

**40**Tech Regular

#

**41**
That is a known Xray problem which still seems to be not sorted out yet.

#

**42**Tech Regular

So speed and radius are related in the case of centrifugal acceleration.

#

**43****Ok then.. for those still interested...**

Sure many factors will apply, as mentioned the engine tune, power band, engagement point and type of track. At a smaller track with tight corners, the inertia required to move the weight of the car on a stop start track may be too harsh.

------------------------------------------

Back to the task at hand...

I think I've come to the conclusion that the weight of the throw and the hardness of the spring or tension applied with preload has more to do with the acceleration of the clutch shoe than the amount of gap. The amount of gap as RBakker said has to do with the forces being applied by the spring to cause slippage. How to measure the differences in forces applied and tension in the spring is my next quest and I'll likely need more help from those of you who like the physics as much as the performance.

This is because the radius change in the throw is negligable and even a 1mm gap will only change the distance on a 30mm clutch (hence velocity) from 94.25 mm per revolution, to 97.38mm per revolution an that's only a change of around 3%. So while the forces do change slightly, they arent significant.

So here goes, I need you keeping check on this RBakker!!

1g throw, 30mm clutch, 10k rpm engage (for one throw only)

At 10k/rpm the velocity of the throws are prior to engagement:

94.25mm * 10,000 / 60 = 15707.96 mm/s or 15.7 m/s

97.38mm * 10,000 / 60 = 16231.56 mm/s or 16.2 m/s

Centrifugal force calcs

15.7^2/.015 * 1g = 16432.6

16.2^2/.0155 * 1g = 16931.6

Again a 3% increase in force.

Increasing centrifugal force.

A Grub screw, a large one may weigh .2 grams. this is a significant factor of up to 20% more force!

*Last edited by blis; 03-11-2015 at 07:38 AM. Reason: Corrected Radius calc*

#

**44**So speed and radius are related in the case of centrifugal acceleration.

#

**45**Tech Regular

It shouldn't be difficult for an engineer if it is your passion to understand the physical world. The more I learn, the more I learn there is more to learn.

The real challenge is perceiving these forces in three dimensions.

1) The Centrifugal / Centripetal (Newton said we cant have one without the other)

2) The pivot on the hinge ( The deformation of the force applied in two directions ) Hinged on the pin and ramp on the flywheel.

3) The Ramp of the flywheel ( The friction, mass, angle )

The real challenge is perceiving these forces in three dimensions.

1) The Centrifugal / Centripetal (Newton said we cant have one without the other)

2) The pivot on the hinge ( The deformation of the force applied in two directions ) Hinged on the pin and ramp on the flywheel.

3) The Ramp of the flywheel ( The friction, mass, angle )

point 1 and 3 are understanded ok, point 2 is the problem i think:

The centrifugal force is always in radial direction of the cranck axis, the value is determined by the mass of fly weight, radius of the center of gravity of the fly weight and the tangential speed of the CoG of the fly weight.