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Old October 17th 20, 06:37 PM posted to rec.aviation.soaring
Tango Eight
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Default Wheel brake effectiveness standards

Let me provide another perspective.

I had around 300 landings in my ASW-20B. I don't immediately recall the number of field landings, but likely 10 ish. That glider, built in 1985, has a 5.00-5 Cleveland wheel / hydraulic disk. That glider weighs as much as a modern 18m glider (no engine). 380 kg dry, with me in it.

My normal habit following a field landing is to walk off the landing roll and self assess. Those landing rolls were never over 250', all but one or two were 200, right on the nose.

So the short answer is: that problem has been solved for 35 years. Copy what works, worry about more important things.

T8

On Saturday, October 17, 2020 at 11:59:10 AM UTC-4, Kenn Sebesta wrote:
On Saturday, October 17, 2020 at 10:29:49 AM UTC-4, Brian wrote:
IMO the aero dynamic forces are mostly not relevant to the brake sizing..

Hmmm, how sure are you on this? I can land my AC-5M at 45kts and roll to a stop within 700' of touchdown. That's some pretty good drag. A lot is obviously coming from the grass, but I suspect with full airbrakes there's a lot of aerodynamic drag as well.
Since you should have a theoretical Drag number and be able calculate about how much drag your airframe and drag devices (flaps and spoilers) produce you could probably calculate a stopping distance with no brake. Probably need to add a small factor for wheel and bearing friction. And then decide how much more energy you want the brake to absorb.

That's a great idea. Where would I get that drag number? It's probably not too different between airplanes so any airplane would probably be typical enough for rough calculations.
But more realistically you want the brake to probably stop you in less than 500-1000 feet depending on the landing speed and weight of the aircraft. The Aerodyamic braking is going drop off of so fast that you can just consider it a design safety factor for the wheel brake system.

How much energy is lost to rolling and air resistance is the critical value to sizing the rotor. As you rightly point out, we can assume the brake rotor system is adiabatic and so all energy goes into heating up the steel. Twice as much steel means half as much temperature rise. So more steel helps keep temps lower but once you're below the critical glazing and warping temperatures extra thermal mass doesn't have any benefit.
That leaves you with just the Kinetic energy of the aircraft at touch down being converted to the amount of the heat the brake can absorb. Again the stopping happens so fast you can pretty much assume none of the heat is going to be dissipated during the braking action. So how much heat does the brake need to be able to absorb?

We can calculate the energy as 1/2 * m* v^2 minus whatever energy is lost to rolling and air resistance. So if we assume 50% for standard landings then we can get away with a 50% smaller disc rotor. That's pretty significant!
The last consideration is the braking force that the brake can generate.. Can it generate enough torque to put the glider up on its nose? You cann make some assumptions about how much elevator forces and resist and or assist this.

Is burying the nose skid more effective than pure wheel braking? I feel like the answer is yes, but it'd be great to have this confirmed.