artL
New member
Oy vey. Not sure I want to go here, but at the risk of kickback I’d like to comment further on the brake discussion, as I think there may be some confusion when it comes to matching the master cylinder to the wheels and brakes. Not regulatory confusion, but the hydraulic system itself.
We have a closed hydraulic system on the Bellanca brakes. In a closed system, pressure is equal throughout. Since fluid is not compressible, once the brake pedals are depressed to the point where the brakes are in contact with the wheels, no further pedal movement is theoretically possible. In effect there is a solid connection between the brake pedal and the brakes and any further force on the brake pedal results in increased braking force at the wheels. The energy required to stop our airplane is based on several factors - most significantly the speed and the weight of the aircraft. The wheels and brakes are sized by the manufacturer to meet those requirements, and as noted in Blimpy's post the Cleveland Wheels and Brakes requires 400 to 500 psi at full gross weight to stop the Bellanca.
This is where it gets interesting. When matching the master cylinder to the wheels and brake requirements, any master cylinder that fits in the airplane and can deliver 400 to 500 psi at the outlet port meets the requirements and is satisfactory.
Here is why.
There are three factors to consider:
Pressure (P) which is the internal system pressure expressed in pounds per square inch (psi);
Force (F) which is the force exerted against the brake pedal and ultimately against the brakes, expressed in pounds;
and Area (A) which is the surface area of the piston (Bore) in the master cylinder or of the brake cylinder at the wheels, expressed in square inches.
The relationship between these factors is expressed in the formulas:
Pressure equals Force divided by Area or P = F/A
so we can solve for the Force as F = P*A
or we can solve for the Area as A = F/P
We know that we need 400-500 psi, therefore let P = 500 psi.
Any combination of Force and Area that will produce 500 psi is theoretically acceptable. Using the example of the Cleveland Master Cylinder given in Blimpy's post, we can look up the P/N 10-35 and see that it has a piston diameter (Bore) of 0.625”.
Using the formula for area of a circle (Area = pi * radius squared) (A = π * rˆ2) we know the area of piston is 0.306 sq. in.
Solving for Force (F = P * A) we get (500 psi * 0.306 sq in) = 153 lbs. The Force required is a measure of how much pressure you can exert on the brake pedals. We need 153 lbs. force to generate 500 psi system pressure. With the same master cylinder (same size piston) you would need 122 lbs. force to generate 400 psi.
This correlates very favorably with a human factors study done in 1971 at Wright-Patterson, where foot forces were measured from 100 pilots of various sizes and weights to determine the affects of brake pedal angles and seat positions on foot force. Neutral pedal forces were around 140 lbs. at a 25˚ pedal angle and extended leg forces were as high as 185 lbs. at 25˚ pedal angle.
The full study is available at http://dviaviation.com/files/38801004.pdf
Since the design of the Bellanca brake system is already established, I would look for a master cylinder that would provide 400 - 500 psi system pressure with a 120 - 140 lb. pedal force.
Using the formula A = F/P you would need a master cylinder with an area between 0.24 sq in (F = 120 lb / P = 500 psi) and 0.35 sq in (F = 140 lb / P = 400 psi).
This means that any master cylinder that fits in the airplane and has a piston diameter (Bore) between 0.552” and 0.667” will do the job. What about the stroke you ask? It isn’t relevant to determining the Pressure since this (our application) isn’t a pressure compensating or power brake system - the stroke only comes into play to seat the brakes against the wheel, and our needs are small.
I think this brief analysis, along with the concepts outlined in AC43.13-1B (Acceptable methods, techniques, and practices aircraft inspection and repair) along with AC 23-27 (Parts & Material Sub for Vintage Aircraft) would form the basis for any proposed STC, 8110-3 or Form 337 to use any master cylinder that fits the need.
Any comments, corrections, questions or challenges are accepted and appreciated.
We have a closed hydraulic system on the Bellanca brakes. In a closed system, pressure is equal throughout. Since fluid is not compressible, once the brake pedals are depressed to the point where the brakes are in contact with the wheels, no further pedal movement is theoretically possible. In effect there is a solid connection between the brake pedal and the brakes and any further force on the brake pedal results in increased braking force at the wheels. The energy required to stop our airplane is based on several factors - most significantly the speed and the weight of the aircraft. The wheels and brakes are sized by the manufacturer to meet those requirements, and as noted in Blimpy's post the Cleveland Wheels and Brakes requires 400 to 500 psi at full gross weight to stop the Bellanca.
This is where it gets interesting. When matching the master cylinder to the wheels and brake requirements, any master cylinder that fits in the airplane and can deliver 400 to 500 psi at the outlet port meets the requirements and is satisfactory.
Here is why.
There are three factors to consider:
Pressure (P) which is the internal system pressure expressed in pounds per square inch (psi);
Force (F) which is the force exerted against the brake pedal and ultimately against the brakes, expressed in pounds;
and Area (A) which is the surface area of the piston (Bore) in the master cylinder or of the brake cylinder at the wheels, expressed in square inches.
The relationship between these factors is expressed in the formulas:
Pressure equals Force divided by Area or P = F/A
so we can solve for the Force as F = P*A
or we can solve for the Area as A = F/P
We know that we need 400-500 psi, therefore let P = 500 psi.
Any combination of Force and Area that will produce 500 psi is theoretically acceptable. Using the example of the Cleveland Master Cylinder given in Blimpy's post, we can look up the P/N 10-35 and see that it has a piston diameter (Bore) of 0.625”.
Using the formula for area of a circle (Area = pi * radius squared) (A = π * rˆ2) we know the area of piston is 0.306 sq. in.
Solving for Force (F = P * A) we get (500 psi * 0.306 sq in) = 153 lbs. The Force required is a measure of how much pressure you can exert on the brake pedals. We need 153 lbs. force to generate 500 psi system pressure. With the same master cylinder (same size piston) you would need 122 lbs. force to generate 400 psi.
This correlates very favorably with a human factors study done in 1971 at Wright-Patterson, where foot forces were measured from 100 pilots of various sizes and weights to determine the affects of brake pedal angles and seat positions on foot force. Neutral pedal forces were around 140 lbs. at a 25˚ pedal angle and extended leg forces were as high as 185 lbs. at 25˚ pedal angle.
The full study is available at http://dviaviation.com/files/38801004.pdf
Since the design of the Bellanca brake system is already established, I would look for a master cylinder that would provide 400 - 500 psi system pressure with a 120 - 140 lb. pedal force.
Using the formula A = F/P you would need a master cylinder with an area between 0.24 sq in (F = 120 lb / P = 500 psi) and 0.35 sq in (F = 140 lb / P = 400 psi).
This means that any master cylinder that fits in the airplane and has a piston diameter (Bore) between 0.552” and 0.667” will do the job. What about the stroke you ask? It isn’t relevant to determining the Pressure since this (our application) isn’t a pressure compensating or power brake system - the stroke only comes into play to seat the brakes against the wheel, and our needs are small.
I think this brief analysis, along with the concepts outlined in AC43.13-1B (Acceptable methods, techniques, and practices aircraft inspection and repair) along with AC 23-27 (Parts & Material Sub for Vintage Aircraft) would form the basis for any proposed STC, 8110-3 or Form 337 to use any master cylinder that fits the need.
Any comments, corrections, questions or challenges are accepted and appreciated.