Drive clutch shift weight????

[recguy]

Does anyone know exactly what this number is reflecting?
As i add weight to the drive clutch this number decreases. Can someone explain?

Thanks

 

[Barker]

This article might help!
[FONT=VERDANA, ARIAL, HELVETICA]In this article, I will attempt to explain how different weights and springs change clutch action. No doubt some of you have been told by a racer or clutch manufacturer: "Youneed these new weights and springs" or "You need the tungsten weight kit". Chances are you've never been told why you need them -- only that you "need" them to go faster (or solve your clutch problems).
Hopefully, after reading this article, you'll have a good idea of how and why these changes to the springs and weights effect clutch action.
In previous magazine articles, I've mentioned how a clutch could theoretically be designed that would be almost totally horsepower insensitive (in other words; slip at almost the same rpm no matter how much or how little power is applied). The mathematics behind this theory are not complicated. To simplify things even more, we will assume that this is a very simple 2-shoe centrifugal clutch (no weights bolted to the shoes, no pivoting shoes). The theory, however, is exactly the same no matter what type of clutch it is, from the most exotic zillion spring, tungsten weight disc axle clutch to the simplest fun-kart clutch. All clutches have some type of mechanism that converts some type of centrifugal force to pressure between fricton surfaces, so the theory will always be the same.
The formula for centrifugal force is: Force = .000341 Wrn²
"W" is equal to the weight of the "shoe" (in lbs.)
"r" is the radius of the center of mass of the shoe (in feet) from the center of the clutch
"n" is the speed of the rotating weight in rpm [which gets squared in the formula]
Or in plain English; The force the shoe will apply is equal to:
.000341 x (weight of shoe) x (radius of the center of mass [in feet]) x rpm squared
Got it ?
Now in order to proceed with this theory without making it overly complicated, we need to make some basic assumptions. Let's assume the following for the sake of simplicity:

• Each shoe needs to exert 100lbs of force (at 10,000) onto the friction surface in order create enough drag to stall the engine at 10,000 rpm. What type of engine we use is not relevant to this example, so we'll just assume that 10,000 is the "right" rpm.

• The radius of the center of mass of the shoe (weight) is at 1.2" -- ( in other words, .1 feet).

• This clutch has NO springs at all, just 2 shoes flying out against a drum.

Now in order to solve for the weight of the shoe, the formula will have to be turned around a bit:
W(weight of shoe) = 100(lbs) divided by [ .000341 x .1(radius) x 100,000,000(rpm²)]
Trust me on this............... the solution will give us the weight that the shoe needs to bein order to apply 100lbs of force to the drum at 10,000rpm. The answer is .0293 lbs. The shoe needs to weigh .0293 lbs in order to apply 100lbs. of force to the friction surface at 10,000rpm.
Actually, it's not necessary to figure out the weight of the shoe to show you what more spring and weight do to clutch action, but knowing how the formula works will give you a better understanding of the following examples.
To show what effect more spring and weight have on a clutch, we first need to calculate what amount of force the shoe (in our imaginary clutch) will exert at say.......... 9900 rpm and 10,100 rpm. (Remember, this clutch has NO springs whatsoever). To figure out how much force the shoe will exert at 9900 rpm, all we have to do is multiply the force (100lbs) by (9900rpm)² and divide by (10,000rpm)².
In other words 100 (lbs) x 98,010,000 / 100,000,000 = 98.01 lbs (of force @ 9900rpm)
And........ 100 (lbs) x 102,010,000 / 100,000,000 = 102.01 lbs (of force @ 10,100rpm)
You can see that the force increases or decreases very gradually with RPM.
Now let's make a RADICAL change to the internals of the clutch:
We add a 500 LB spring to each shoe (in other words, a spring that is trying to pull the shoe toward the center of the clutch) ............ so in order for the shoe to still exert 100lbs of force to the friction surface at 10,000 rpm, the shoe will actually have to apply 600lbs of force (of which 500lbs is absorbed by the spring). So the shoe needs to weigh SIX TIMES as much in order to accomplish that. The shoe now weighs .1758lbs (it used to weigh .0293lbs which we have multiplied by six).
Our shoe now weighs 600 lbs at 10,000rpm (remember, 500lbs is carried by the spring)
So let's go back to solving this for 9900 and 10,100 rpm.
So.... 600 x 98,010,000 / 100,000,000 = 588.06 lbs (- 500lbs by spring) = 88.06 lbs !!
And ... 600 x 102,010,000 / 100,000,000 = 612.06 lbs (- 500lbs by spring) = 112.06 lbs !!
Now look at the 2 examples ...... with no springs, the force the shoe exerted on the friction surface at 10,100 only increased a bit over 2 % from 10,000 .... but with a bunch of spring and weight in the clutch, the force increased over 12 % !!!!
Conversely, the force fell off by less than 2 % in the first example, but fell off by almost 12% in the second example.
As you can imagine, by carrying this to the extreme, you could theoretically build a clutch that would allow a Yamaha to slip to 10,000rpm, but you couldn't make it slip 10,500 with a 100 HP !!! The force on the friction surfaces would increase so radically with only a minor RPM increase, it would be virtually impossible for the clutch to "overslip".
Conversely, you could reduce the power of the motor by 50 %, and the "slip" rpm would still be almost 10,000 because the force of the shoe would "drop-off" so rapidly below 10,000rpm. Remember that the spring force never changes (no matter what the RPM, but the centrifugal force increases or decreases at the SQUARE of the speed).
Now back to the real world ................ (where things aren't so mathematical):
No clutch manufacturer has ever figured out how to make the weights in your clutch 5 times as heavy, or how to incorporate an enormous spring into the clutch mechanism. However ....... the theory that is illustrated above, shows WHY there has been a slow but steady effort over the years to make the "weight" heavier, and the "springs" stronger in clutches. The more spring and weight in a clutch, the more the clutch tries to "hold" a given RPM regardless of how much or how little power is applied.
Increasing the weights and springs by only 50% can make a noticeable difference in clutch action.
[/FONT][FONT=VERDANA, ARIAL, HELVETICA]Unfortunately, this can (and does) lead to a Catch-22 situation: Making the clutch a bit better leads to the pipe manufacturers building slightly more radical exhaust pipes, which leads to the clutch guys trying to make the clutch better, which leads to the pipe guys trying to ...... YOU GET THE PICTURE....... !!![/FONT]

 

[recguy]

This all changes with elevation increase too. If i keep the same spring and add weight and decrease elevation the numbers remain close to the same until a certain weight then it just changes drastically.

Let the engineers do the R&D, i was just trying to figure it out.

 

[Barker]

This all changes with elevation increase too. If i keep the same spring and add weight and decrease elevation the numbers remain close to the same until a certain weight then it just changes drastically.

Let the engineers do the R&D, i was just trying to figure it out.

i put all that on there and now your just going to let the engineers figure it out! Just because they have a big ring! Come on! Lol!