Drum vs Disk swept area calculation

[zblu]

Anybody have an idea/formula on how to work this so as to come up with some parameters, ie is it the drum pad area to use on one side of the calc and the rotor or disk pad area on the other side or?

 

[Dave Russell]

Not sure about the question. On a disc brake, the swept area would be the disc area that was in contact with the pads. If the disc pad surface were 11" outside diameter & 8" inside contact diameter, the swept area would be the area of the 11" circle minus the area of the 8" inside circle times two. Or in this case, 95 square inches minus 50 square inches = 45 sq inches times two sides or = 90 square inches per wheel swept area.

((11" x 11") X .7854) = 95 sq inches, minus ((8" x 8") x .7854) = 50 sq inches, & 95 - 50 = 45 per side of disc & 45 x 2 sides = 90 sq inches per wheel swept area.


For a drum with an 11 inch inside diameter & 2.25" wide, the swept area would be 78 square inches per wheel. (11" x 3.14) X 2.25" = 77.7)

Is this what you are after or am I confused?
D

 

[zblu]

Thanks Dave,
What I was looking at was the area of contact as well, so as to be able to set some limits re putting disks on, so from memory the drum pads would have approx 50 sq inches of contact area and the pads on the disks would have say 7"x2= 14 sq inches, am ignoring application pressures at the moment, just wondering if there is a chart/correlation somewhere to work with

 

[nevets]

Dave
perhaps you can shed some light on an ongoing debate I am having with one of our engineers. Accoding to my engineer, the amount of swept area does not increase braking force...he claims that no matter how many sq inches of contact surface, 10 pounds of force (clamping pressure) is still just 10 pounds of force, and that the real benefit of distributing the force over a larger swept area is in cooling effiency. Intuitively, this does not sound right to me. Any thoughts?

 

[Dave Russell]

I tend to agree with the guy. If a given contact force is spread out over a greater surface area, the psi goes down, the total force remains the same.

Friction is pretty much dependent on coefficient of friction of the materials & the clamping pressure & independent of friction surface area.

A larger diameter disc or drum does give the brake more mechanical leverage on the wheel & would require less friction, clamping pressure, to get the same stopping power. A smaller disc diameter would require more pressure. Also a larger swept area gives more cooling & less wear.

In case you are wondering, this does not apply to tire grip. Soft tires actually have a good measure of adhesion in addition to friction, & can benefit from more surface area in contact with the road. Wet or icy surfaces, where there is no adhesion, excepted.
D

 

[Dave Russell]

[ QUOTE ]
Thanks Dave,
What I was looking at was the area of contact as well, so as to be able to set some limits re putting disks on, so from memory the drum pads would have approx 50 sq inches of contact area and the pads on the disks would have say 7"x2= 14 sq inches, am ignoring application pressures at the moment, just wondering if there is a chart/correlation somewhere to work with

[/ QUOTE ]
See above. Swept area & contact area have little effect on braking power. It indirectly affects mechanical advantage, cooling, & component life. I don't know of any charts. I think it's best to go with something that is known to work for your application.
D

 

[Keoke]

Nevets, /ubbthreads/images/graemlins/iagree.gif wid that engineer fella too.---Keoke- /ubbthreads/images/graemlins/laugh.gif