For your reading pleasure:
{This document "cleaned up" a bit in July, 2000}
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: "You
need 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 for a kart 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 (or even a clutch on a Top Fuel dragster, for that matter) to the simplest fun-kart clutch. All clutches have some type of mechanism that converts some type of centrifugal force to
pressure between friction 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 pounds)
"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 [to the friction surface] is equal to:
.000341 x (weight of shoe) x (radius of the center of mass [in feet]) x (rpm squared)
In order to proceed with this theory (without making it overly complicated), we should make some
basic assumptions:
Let's assume the following for the sake of simplicity:
- Each shoe needs to exert 100 pounds of force (at 10,000 rpm) onto the friction surface in order create enough drag to stall the engine at 10,000 rpm. (By "stall", I mean that if you hold the brake, and give the engine full throttle, the engine will notrev any higher than 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 center of mass of the shoe is at 1.2" radius -- (in other words: .1 feet). Note: This means 1.2" from the centerline of the crankshaft... in other words: from the center of rotation of the entire assembly.
- This clutch has NO springs at all, just 2 shoes flying out against the bore (inside diameter) of a drum.
(once again: it makes absolutely no difference if the clutch is a 12-finger, 5-disc Top Fuel dragster clutch, or a 2 shoe centrifugal clutch on a mini-bike... friction is friction, force is force... nothing changes except the size of the numbers).
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 (pounds) 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 be in order to apply 100 pounds of force to the friction surface at 10,000 rpm. The answer is .0293 lbs. The shoe
needs to weigh .0293 lbs (on a scale) in order to apply 100 pounds of force to the friction surface at 10,000 rpm.
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.
Now to the good stuff...
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. This will give us a good idea of how rapidly the force increases with rpm... or how rapidly it
decreases with rpm. (Remember, this clutch has NO springs whatsoever).
To calculate how much force the shoe will exert at 9900 rpm, all we have to do is multiply the force (100 pounds) by (9900 rpm)² and divide by (10,000 rpm)².
In other words 100 (pounds) x 98,010,000 / 100,000,000 =
98.01 pounds (of force @ 9900 rpm)
And... 100 (pounds) x 102,010,000 / 100,000,000 =
102.01 pounds (of force @ 10,100 rpm)
We now have a good idea of how this imaginary clutch "acts" above and below the stall rpm.
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 pound 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 100 pounds of force
to the friction surface at 10,000 rpm, the shoe will actually have to apply 600 pounds of force (of which 500 pounds is overcome/balanced by the spring). So the shoe needs to weigh SIX TIMES as much in order to accomplish that. The shoe now weighs .1758 pounds (it used to weigh .0293 pounds which we have multiplied by six).
Our shoe now "
weighs" 600 pounds at 10,000 rpm, but just like before, it applies 100 pounds of force to the friction surface. (remember: 500 pounds is balanced by the spring)
Now 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 (minus 500 pounds being balanced by the spring) =
88.06 pounds !
And ... 600 x 102,010,000 / 100,000,000 = 612.06 lbs (minus 500 pounds being balanced by the spring) =
112.06 pounds !
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 "stall" a 15hp Yamaha engine at 10,000 rpm, but you couldn't make it "stall" at 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 "stall" rpm would still be almost 10,000 because the force of the shoe would "drop-off" so rapidly below 10,000 rpm. 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 go kart engine 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.
Unfortunately, in karting 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 -- !
Additional notes (and things to think about):
The theory explained in this article... while significant... is by no means the only thing that effects how a clutch works.
Other things that contribute to clutch "action":
- Coefficient of friction of the friction surface(s)
- Changes in coefficient of friction with temperature
- Static vs. dynamic coefficient of friction
- Rigidity of the clutch assembly
- Geometry of the "lever" (disc clutches)
- Radius of the contact point of the lever (disc clutches)
- Static clearance of the disc stack (disc clutches)
- etc. etc.
Pete Muller
