8" pad vs. 6.5" pad

[ahains]

Hi Zet,
I'm glad to see I'm not the only geek that likes to pull the math out on hobby problems

I was actually thinking the same way yesterday/today, and wondering how much faster you need to spin an X sized pad compared to a Y sized pad, to achieve the same level of work.

All things being equal, I think the amount of work will essentially be defined by area_of_foam * distance_of_foam per revolution.

My calculus is rusty, but I think it would solve this well. I'll post my thoughts in case some one can do the integral for me

For the area, we look at the area of x and x+dx like:
dx area = outerArea - innerArea
outerArea = x+dx
innerArea = x
so this is ((x + dx)^2 * pi) - (x^2 * pi)

distance per revolution is the average of outerCircumference and innerCircumference (add together and divide by 2).
outerCircumference = 2*(x+dx)*pi
innerCircumference = 2*x*pi
so this is (x+dx)*pi + x*pi

Then you multiply this area and distance:
(((x+dx)^2)*pi - (x^2)*pi) * ((x+dx)*pi + x*pi)

Simpify a tad:
pi^2 * ((x+dx)^2 - x^2)*(2b+dx))

So I'm not sure what the antiderivative is of that, but I think if you apply it over 0 to 6.5 and 0 to 8, you'll essentially get the difference in work between the two pad sizes.
Of course this assumes a linear amount of work as speed increases (i.e. one square inch of foam going two inches in one second is twice as effective as it going one inch), and I'm not sure this is correct.

Anyone remember how to do antiderivatives? You might want to double check my math too..

[ahains]

Probably I'm just spewing incorrect pseudo math, but I think it is simpler than I was thinking.
Obviously the amount of work for inches 0 through 6.5 are the same on both pads. So I think we can just calculate the work (area * distance) for both pad sizes.

circumference(6.5) = 20.41, c_averagy(0 to 6.5) is 10.2
c(8) = 25.12, so c_average(8) is 12.56
c_average(6.5 to 8) = 25.12+20.41 / 2 = 22.765"

a(6.5) = 33.16625
a(8) = 50.24
area of (6.5 to 8) = 50.24 - 33.16625 = 17.07375

a(6.5) * avg_c(6.5) = 33.16625 * 10.2 = 338.29575 work
a(6.5 to 8) * avg_c(6.5 to 8) = 17.07375 * 22.765 = 388.68391875 work
a(8) * c_average(8) = 50.24 * 12.56 = 631.0144

I think I probably have an error here, because those last two values of work should add up to the work of the 8" pad but they don't
In any case, my guesstimate is that an 8" pad does around twice as much work at a given RPM as a 6.5" pad. It's true that the maximum speed at the outer edge is only about 25% faster, but this is pretty significant since there is about 50% more pad area AT this faster than the 6" pad speed.

[Zet]

Wow.. those are some fancy formulas

While there is about 50% more pad area on the 8" pad, only the outer edge would be spinning 25% faster, so across the area it would be spinning from 0 - 25% faster, than the outer edge of the 6.5" pad

And there is another factor we have not taken into consideration yet, which is the pressure applied to the pad And this gets even more complicated, as the pressure would be highest directly under the backing plate, as the edges of the pads tend to bend upwards when pressure is applied. This would be more so for the 8" pads. Equally important is pressure applied per square inch.

Wow, this is getting complicated. Maybe we need a math forum on here, where we could solve various detailing problems