Mixing Roller Weights

[Vivo]

Do or have any of you guys did this? I've seen and read people mixing roller weights... I need to hear from your experiences...   

Here's some experiment with the Super 8 (not mine)

1000rpm center spring 12gx3/13gx3   acceleration and top end
1000rpm center spring 13gx6              acceleration, mid, and top end
1000rpm center spring 13gx3/14gx3   mid and top end
1000rpm center spring 13gx3/15gx3   mid and top end

1500rpm center spring 12gx3/13gx3   acceration and mid
1500rpm center spring 13gx6              acceleration, mid, and top end
1500rpm center spring 13gx3/14gx3   mid, top end
1500rpm center spring 13gx3/15gx3   mid, top end

[Vivo]

Well, I guess no one here does this....

[MotoRandy123]

It's not really recommended as you only have 3 doing the work. It will rev higher as the weight of
just the 3 is what you get. Some people do it to see if lower would be better...

[blue]

Right now I have 3 5.5 and 3 6 in my scoot only because I ran out of rollers. so far so good take off is A little faster and the top ends takes a little longer to get there but Im ok with that.

[trninka]

As I know, you can mix it but spread them properly. I am willing to mix on my agility 3x 7gr and 3x 8gr, by the way: 1x 7gr, 1x 8gr, 1x 7gr and so on...

It is not recommended to mix the rollers that has too much different in weights, for example 8gr with 5 gr...

[ScooterCommuter]

It's not really recommended as you only have 3 doing the work. It will rev higher as the weight of
just the 3 is what you get. Some people do it to see if lower would be better...

Actually with the physics involved, that's not true. Every roller or slider is exerting an outward force proportional to its mass. The sum of those forces is converted to an axial force by the ramps in the variator and acts against the spring. The heavier rollers are contributing a greater amount to that sum, to be sure, and over time would wear faster than the lighter ones but every rollers contribution to that sum is the same irrespective of what the weights of the other rollers are.  You would see exactly the same performance with 6x5.5g as with 2x5.0g, 2x5.5g, 2x6.0g. Only the rate at which the rollers wore down and the wear on the ramps would differ.

[MotoRandy123]

That just doesn't sound right to me. If the weights are close you can ignore it a bit but
if I had a 5 pound weight and a 1 oz weight and spun them around I assume the 5 pounder
will be doing all the work... Mass x acceleration the masses are different so the forces are different...

[LoveMyKymco]

Were talking centrifugal force or inertial force here.... Total inertial force is dictated by total mass which then leads to centrifugal force.
If I can figure out the commands for the symbols I can make the equation. Explaining it takes more typing than I feel like doing at the moment.

[LoveMyKymco]

And I mix my weights all the time to get the desired total weight. Roller weights are dictated by the riders weight and driving style, and cant be equated into a set formula unfortunately. At least not easily.

[ScooterCommuter]

That just doesn't sound right to me. If the weights are close you can ignore it a bit but
if I had a 5 pound weight and a 1 oz weight and spunt them around I assume the 5 pounder
will be doing all the work... Mass x acceleration the masses are different so the forces are different...

Consider the rollers individually. Centripetal force is (mv^2)/r. So think of a roller in 5k rpm motion along a 3cm (0.03m) radius circle. The velocity of that roller is (2pi*r)/time and time is 1/5000 so that velocity is (2pi*0.03)/(1/5000) = 942 m/s. Plugging that into the force calculation, an individual 5.5g roller is trying to move outward against the ramps with a force of (0.055^2)/0.03 = 1626834 newtons. 6 of those will apply a force to the variator of 9761 kN.

If we repeat the same calc with 6.0 and 5.0 masses, an individual 6.0g roller applies a force of 1774728 N, and a 5.0g roller one of 1478940. (2*1478940)+(2*1626834)+(2*1774728 ) = exactly the same result, 9761 kN.

[ScooterCommuter]

Or making even simpler,  the rigid construction of the variator assembly means that to use the swung weight analogy you are assuming all the weights are tied to the same string. If you swing 6 3lb weights on one string you are swinging 18 lbs. Does it take any less or more effort to swing a string to which you have tied 2 2lb weights, 2 3lb weights and 2 4lb weights?

[ScooterCommuter]

Oops I goofed my calc shouldnt have been 1/5000 but instead 60/5000. So that means my numbers were off by factor of 60.

[Vivo]

Consider the rollers individually. Centripetal force is (mv^2)/r. So think of a roller in 5k rpm motion along a 3cm (0.03m) radius circle. The velocity of that roller is (2pi*r)/time and time is 1/5000 so that velocity is (2pi*0.03)/(1/5000) = 942 m/s. Plugging that into the force calculation, an individual 5.5g roller is trying to move outward against the ramps with a force of (0.055^2)/0.03 = 1626834 newtons. 6 of those will apply a force to the variator of 9761 kN.

If we repeat the same calc with 6.0 and 5.0 masses, an individual 6.0g roller applies a force of 1774728 N, and a 5.0g roller one of 1478940. (2*1478940)+(2*1626834)+(2*1774728 ) = exactly the same result, 9761 kN.

Nice!     my nose bleeds....

[ScooterCommuter]

ok between the typos from sending from my phone and goofing one number I cant let that stand so I'll repeat it correctly..

Tangential velocity of a single rotor at 5k rpm is (2pi*r)/t so with r=0.03m and t=60/5000 we get v=15.71 m/s

Force exerted by that single roller is (m*v^2)/r, so if v^2 = 15.71*15.71 = 246.74, then (v^2)/r = 246.74/0.03 = 8224.67, and force exerted by a roller of mass m is m*8224.67.

For 5.0g (0.05 kg) rollers force = 411.23 N each
For 5.5g rollers force = 453.36 N each
For 6.0g rollers force = 449.48 N each

Therefore comparing roller sets:
6x5.5g = 2.714 kN
2x5.0g+2x5.5g+2x6.0g = 2.714 kN

So even with my typos and a math goof it still works. If total weight is equal you still get equal force acting against the spring.

(Edited to add: I performed the calcs at full precision in windows calculator to avoid rounding errors as much as possible, but typed the intermediate steps rounded to 2 or 3 places because I didn't feel like subjecting you to huge decimals. That's why some of those steps look off in the least significant digits.  Yeah I admit to being a math geek. Sue me.  )

[MotoRandy123]

I guess the answer I needed was that due to the rollers being the same diameter the forces all
add up even though some are less and some are more. If the driven surface could tilt and the
masses were not evenly spaced then the variation in mass wouldn't work (like the equivalent lesser
mass) ...