Topic: Can somebody explain this? Drilling a rotor increases it's surface area.

[fastbroshi]

I assume they meant pad swept area?  If that is what they meant, then how?  From the webpage:  http://www.vintagebrake.com/tips.htm

Sixth heading under Disk Brake Tips

Disc lightening--drilling vs. thinning: both will reduce mass and therefore heat sink capability. Some pluses for drilling.: If the holes drilled are smaller in diameter than the thickness of the disc, surface area is increased.

[Hopper]

I think what they mean is that the total surface area of the disc is increased, therefore increasing the cooling area, but not increasing the swept area.

The surface area of the cylinder that is the inside surface of the hole is greater than the former two small circles of flat space that were sacrificed by drilling the hole.

[fastbroshi]

Okay, but do you have the formula to figure that out, jk

[Markcb750]

(Pi x D) x L x N

Pi =3.14159....
D=drill dia
L= thickness of rotor
N= number of holes


Opps, what can say I'm old.

[fastbroshi]

Nice, thanks Mark.

[turboguzzi]

its also well explained in john robinson's book "motorcycle tuning - chassis"

the crazy drilling fad (more holes than disc... ) was really massive when discs where thick and heavy due to novelty/playing it safe.

if the tracks your race on have long straights with heavy braking at the end, you would want to maintain some heat sink capacity and not reduce weight at all costs,

Personally, i run gs750 discs thinned to  4.5 with not much drilling, got holes there for good pad cleaning and wet braking rather than reducing weight,



TG

[mlinder]

(Pi x r^2) x L x N

Pi =3.14159....
r==1/2 drill dia
L= thickness of rotor
N= number of holes

Sorcery! Burn him at the stake!

[crazypj]

To increase the area for cooling as a rule of thumb, the holes have to be smaller than the disc thickness

[fastbroshi]

(Pi x r^2) x L x N

Pi =3.14159....
r==1/2 drill dia
L= thickness of rotor
N= number of holes

Shouldn't it be (2 pi r)* h * number of holes?

http://www.math.com/tables/geometry/surfareas.htm#cylinder

[scottly]

(Pi x r^2) x L x N

Pi =3.14159....
r==1/2 drill dia
L= thickness of rotor
N= number of holes

Oops! This formula gives the volume of the holes, not the surface area! Still useful, regarding drilling discs, with regards to weight removed. Multiply the result by .290, which is the weight in pounds of one cubic inch of stainless steel.

The surface area inside the hole is pi x dia x L

The surface area of the flat area of the disk, were it not drilled, is (pi x r^2) x 2

Break out your calculators, class, and see if the holes have to be smaller than the disc thickness to increase the surface area.  (I measured a couple of stock 750 discs: .270")  

[fastbroshi]

So trying to find the volume assuming a 4mm hole and a 5mm disk thickness, I get 62.83mm cubed.  Multiplied by 0.29, I got 18.22.  is that ounces?  Am I doing that right? 

[scottly]

Convert 62.83 mm cubed to cubic inches, then multiply by .29 to get the weight.
To make things easier, do your calculations in inches in the first place; then you won't have to do the conversion from metric.

[fastbroshi]

DAMMIT!!  Thanks Scott, right in front of me!  So with 100 4mm holes, your weight savings would be .717 lbs or 11.4oz, not too shabby.  Not to mention some extra cooling.

[scottly]

I'll have to check the math, but I'm guessing that is an optimistic weight loss for 100 4mm holes..

[scottly]

I came up with .15243 pounds for 100, 4mm dia holes, in a .270 thick disc.

[fastbroshi]

Can you show your math because I just did it again and came up with something different. 

[crazypj]

I drilled my Yamaha XS650 discs and 'lost' a bit over two pounds  (weighed them before and after drilling)
They are probably the heaviest discs ever fitted to a motorcycle though
 Use half inch holes, but, I could afford to loose some surface area as I fitted dual disc and 4 piston calipers.
 Twin disc models have smaller diameter discs, I used two 'single disc' which are larger dia.

[scottly]

ex: 100- .250" holes, in .270 thick disk:
.125 x .125 = .01565, x3.14 = .0490625, x .270 = .0132468, x 100 = 1.32468 x .290 = .38415 pounds.

pj, depending on the thickness of the discs, the break-even point between internal surface area for cooling, and the swept area is about .5" holes.

[fastbroshi]

Ok I've got it Scott, ty.

[scottly]

Hey, sorry for the hi-jack, but are you interested in the up-coming rally? 10 miles, 100,?

[fastbroshi]

I'll have to read over the thread again for details on when it goes through Texas, but most likely.  Hell, Texas alone is quite a ride, at least 12 hours if you don't stop for nada.

[lone*X]

I'll have to read over the thread again for details on when it goes through Texas, but most likely.  Hell, Texas alone is quite a ride, at least 12 hours if you don't stop for nada.

Orange to El Paso is 856 miles.  At the speed limits on I10 that is 13 hrs 26 minutes.  Course someone could cheat and cut across a corner. 

[scottly]

lone*X, welcome aboard! Check in here:
http://forums.sohc4.net/index.php?topic=81307.0

[gsnorcal]

I'm sure this will be the most useless post I ever made on the interwebs... (and I hope I'm correct in my formulas)

Surface area of an undrilled disc is:

pi*R^2 - pi*r^2 + 2*pi*R*h + 2*pi*r*h

Where:

R is the outer radius
r is the inner radius
h is the thickness

If you drill it, you loose area on the flat surface if the disc:

- n*pi*q^2

but gain the cylindrical walls of the drilled holes:

+ n*2*pi*q*h

where:

n=# holes
q=radius of holes

So, the complete formula for a disc's surface area is:

pi*R^2 - pi*r^2 + 2*pi*R*h + 2*pi*r*h - n*pi*q^2 + n*2*pi*q*h

or

surface of disc - center of disc + outer column + inner column - drilled surface + drilled column

Whew!

[Reference: http://math.com/tables/geometry/surfareas.htm#cylinder]

[754]

if you are ding nominal like 1/2 or 3/8 holes, you can look up the weight(per foot) for steel bar, then x thickness of rotor x # of holes.. to figure weight..