Disc brake rotor question

Joe Riel said:

jim beam said:

Joe Riel said:

How will removing material reduce the operating temperature
or rate of heat dissipation?

Quoted message said:

surface area/mass. it's a two-edged sword, because low mass heats
quicker, but the other side of it means increasing surface area allows
air flow to carry more heat away more quickly.

For a circular hole, unless the radius is less than the thickness
of the material, the drilled material has "less" surface area.

Exactly correct, if you drill a hole of radius r, you remove 2*pi*r*r of
surface area from the disk, but "give back" 2*pi*r*h as the lateral
surface area of the hole. So you reduce the surface area if r > h.

But I think the surface area to volume ratio will always increase with
cross-drilling (see my other post where I went on about cookie cutters).

Ben C said:

Joe Riel said:

jim beam said:

Joe Riel wrote:
> How will removing material reduce the operating temperature
> or rate of heat dissipation?

Quoted message said:

surface area/mass. it's a two-edged sword, because low mass heats
quicker, but the other side of it means increasing surface area allows
air flow to carry more heat away more quickly.

For a circular hole, unless the radius is less than the thickness
of the material, the drilled material has "less" surface area.

Exactly correct, if you drill a hole of radius r, you remove 2*pi*r*r of
surface area from the disk, but "give back" 2*pi*r*h as the lateral
surface area of the hole. So you reduce the surface area if r > h.

But I think the surface area to volume ratio will always increase with
cross-drilling (see my other post where I went on about cookie cutters).

No, in fact it decreases. Consider a ring with

Ro = outer radius
Ri = inner radius
t = material thickness
Rh = hole radius
n = number of holes

Then

V = pi*t*(Ro^2-Ri^2-n*Rh^2)
A = pi*(Ro^2-Ri^2-2*n*Rh^2) + 2*pi*(Ro+Ri+n*Rh)*t

so

h*A/V = 1 + (2*t*(Ro+Ri) + n*Rh*(2*t - Rh))/(Ro^2 - Ri^2 - n*Rh^2)

Let Rh >> 2*t, then we can approximate this by

h*A/V = 1 + (2*t*(Ro+Ri) - n*Rh^2)/(Ro^2 - Ri^2 - n*Rh^2)

Leting
Ae = 2*t*(Ro+Ri)
Ac = Ro^2 - Ri^2
Ah = n*Rh^2
we have

h*A/V = 1 + (Ae - Ah)/(Ac - Ah)

Take derivative with respect to Ah,

d(h*A/V)/dAh = (Ae - Ac)/(Ac - Ah)^2

Because Ae < Ac this derivative is less than zero.
So as the volume of holes increases, the ratio h*A/V decreases.

--
Joe Riel

Joe Riel said:

Ben C said:

On 2006-12-17, Joe Riel <[email hidden]> wrote:

[snip]

Quoted message said:

Quoted message said:

Quoted message said:

For a circular hole, unless the radius is less than the thickness
of the material, the drilled material has "less" surface area.

Exactly correct, if you drill a hole of radius r, you remove 2*pi*r*r of
surface area from the disk, but "give back" 2*pi*r*h as the lateral
surface area of the hole. So you reduce the surface area if r > h.

But I think the surface area to volume ratio will always increase with
cross-drilling (see my other post where I went on about cookie cutters).

No, in fact it decreases. Consider a ring with

Ro = outer radius
Ri = inner radius
t = material thickness
Rh = hole radius
n = number of holes

Then

V = pi*t*(Ro^2-Ri^2-n*Rh^2)
A = pi*(Ro^2-Ri^2-2*n*Rh^2) + 2*pi*(Ro+Ri+n*Rh)*t

I mostly agree with this part (but see below), but if we try these
formulas with some numbers, we can see that A/V increases with the
number of holes. Suppose Ro is 10, Ri is 2, t is 1 and Rh is 0.1, here's
what I get for a range of numbers of holes:

0 holes, volume 301.593, area 376.991, area / volume 1.250
10 holes, volume 301.279, area 382.646, area / volume 1.270
20 holes, volume 300.965, area 388.301, area / volume 1.290
30 holes, volume 300.650, area 393.956, area / volume 1.310
40 holes, volume 300.336, area 399.611, area / volume 1.331
50 holes, volume 300.022, area 405.265, area / volume 1.351
60 holes, volume 299.708, area 410.920, area / volume 1.371
70 holes, volume 299.394, area 416.575, area / volume 1.391
80 holes, volume 299.080, area 422.230, area / volume 1.412
90 holes, volume 298.765, area 427.885, area / volume 1.432

I worked these out by pasting your exact formulas into a computer
program. If you plot these on a graph, you get a straight line with A/V
increasing linearly with the number of holes.

I have one issue with your formula for area, which is that the first
term only counts the undrilled area for one side of the disk, but seems
to count the holes on both sides. In other words, it should be:

2*pi*(Ro^2 - Ri^2 - n*Rh^2)

Using this formula instead produces a similar linear increase in A/V for
n:

0 holes, volume 301.593, area 678.584, area / volume 2.250
10 holes, volume 301.279, area 684.239, area / volume 2.271
20 holes, volume 300.965, area 689.894, area / volume 2.292
30 holes, volume 300.650, area 695.549, area / volume 2.313
40 holes, volume 300.336, area 701.203, area / volume 2.335
50 holes, volume 300.022, area 706.858, area / volume 2.356
60 holes, volume 299.708, area 712.513, area / volume 2.377
70 holes, volume 299.394, area 718.168, area / volume 2.399
80 holes, volume 299.080, area 723.823, area / volume 2.420
90 holes, volume 298.765, area 729.478, area / volume 2.442

Quoted message said:

so

h*A/V = 1 + (2*t*(Ro+Ri) + n*Rh*(2*t - Rh))/(Ro^2 - Ri^2 - n*Rh^2)

What's h?

[snip]

You've lost me with the rest of this!

Here is some simpler math. Suppose it's just a complete disk (never mind
about Ri in other words).

A = 2*pi*(Ro^2 - n*Rh^2) + X + Y
V = pi*t*(Ro^2 - n*Rh^2)

X is an extra term for the additional surface area inside the holes It's
actually (2*pi*n*Rh*t), but it doesn't matter, so long as it's positive
and proportional to n.

Y is an extra term for the lateral area of the disk itself (actually
2*pi*Ro*t).

Now, if it weren't for X and Y, A/V would be 2/t. It wouldn't depend on
n. So, apart from the surface area of the insides of the holes, drilling
a cylinder keeps its A/V ratio the same.

This is explained qualitatively by the "cookie cutter" reasoning-- if a
3D shape is an extrusion, then cutting prisms out of it along its axis
of extrusion will reduce its area and volume by the same amounts. This
is because the thickness is the same everywhere and the volume of the
piece you cut out is given by the product of its area and the thickness.

Y is constant, but when we consider the term X, we have a term that
increases A/V proportionally to the number of holes.

Ben C said:

Joe Riel said:

Ben C said:

On 2006-12-17, Joe Riel <[email hidden]> wrote:

[snip]

Quoted message said:

Quoted message said:

> For a circular hole, unless the radius is less than the thickness
> of the material, the drilled material has "less" surface area.

Exactly correct, if you drill a hole of radius r, you remove 2*pi*r*r of
surface area from the disk, but "give back" 2*pi*r*h as the lateral
surface area of the hole. So you reduce the surface area if r > h.

But I think the surface area to volume ratio will always increase with
cross-drilling (see my other post where I went on about cookie cutters).

No, in fact it decreases. Consider a ring with

Ro = outer radius
Ri = inner radius
t = material thickness
Rh = hole radius
n = number of holes

Then

V = pi*t*(Ro^2-Ri^2-n*Rh^2)
A = pi*(Ro^2-Ri^2-2*n*Rh^2) + 2*pi*(Ro+Ri+n*Rh)*t

I mostly agree with this part (but see below), but if we try these
formulas with some numbers, we can see that A/V increases with the
number of holes. Suppose Ro is 10, Ri is 2, t is 1 and Rh is 0.1, here's
what I get for a range of numbers of holes:

Quoted message said:

I have one issue with your formula for area, which is that the first
term only counts the undrilled area for one side of the disk, but seems
to count the holes on both sides. In other words, it should be:

2*pi*(Ro^2 - Ri^2 - n*Rh^2)

Arrggh! You are correct. Thanks for catching my error.
My computation, alas, falls apart once that correction is made.

Computing d(A/V)/dn at n=0 we get

d(A/V)/dn|(n=0) = 2*(Ro-Ri+Rh)*Rh/(Ro-Ri)/(Ro^2-Ri^2)

for Rh << Ro,Ri, this simplifies to

2*Rh/(Ro^2-Ri^2)

which is always positive.

Quoted message said:

What's h?

I had originally used h for the thickness, than changed it to
t to avoid a conflict with Rh, but didn't change the term on
the left side.

--
Joe Riel