Brakes
- Thread starter A BELLAND
- Start date
irishfield
Registered User
Penetanguishene, Ontario Canada
Still the same airplane weight.... and the wheel will be now turning slower for the same MPH ! Why wouldn't it?
...and I'd rather have minimal brakes in a tail dragger than maximum.
...and I'd rather have minimal brakes in a tail dragger than maximum.
Bob Breeden
Registered User
Alaska
Aurele,
For what it is worth, I have single puck Clevelands on my C-180, and 8.00 x 6.00 Goodyears. When I bought the plane, I had my doubts about the braking strength. But believe me, there is plenty of braking - even with 4 adults, fuel, and a strong crosswind, on landing there is always enough brake on the downwind wheel to keep the big long tail where it is supposed to be. There you have a heavier plane with more energy to dissipate through braking.
The contrast is that you are turning a larger radius tire. 26 inch tires - you can do the math - 3.14159 x your actual tire height gets the circumference. I haven't measured my 8.00 x 6.00's for your request, but I guess they are about 14 inches tall (I'll check soon). 14 x pi is 44 inch circumference. The 26's that I used to run had about a 24 inch height, for a 75.4 inch circumference. So that is about a 2 to 1 ratio - in favor of smaller diameter tire.
Taking your lighter Cub with the single puck brakes I wouldn't hesitate on 26 inch or smaller tires.
Bob Breeden
www.AlaskaAirpark.com
For what it is worth, I have single puck Clevelands on my C-180, and 8.00 x 6.00 Goodyears. When I bought the plane, I had my doubts about the braking strength. But believe me, there is plenty of braking - even with 4 adults, fuel, and a strong crosswind, on landing there is always enough brake on the downwind wheel to keep the big long tail where it is supposed to be. There you have a heavier plane with more energy to dissipate through braking.
The contrast is that you are turning a larger radius tire. 26 inch tires - you can do the math - 3.14159 x your actual tire height gets the circumference. I haven't measured my 8.00 x 6.00's for your request, but I guess they are about 14 inches tall (I'll check soon). 14 x pi is 44 inch circumference. The 26's that I used to run had about a 24 inch height, for a 75.4 inch circumference. So that is about a 2 to 1 ratio - in favor of smaller diameter tire.
Taking your lighter Cub with the single puck brakes I wouldn't hesitate on 26 inch or smaller tires.
Bob Breeden
www.AlaskaAirpark.com
Braking effectiveness is inversely related to tire height..... bigger the tire, less braking.
Tire circumference has little to do with brake effectiveness. Weight being the same, brake effectiveness only relates to tire footprint and the distance the tire/surface contact is from the braking device..... and the brakes ability (hydraulic pressure) to stop the wheel from turning.
A single puck Cleveland brake (especially with stock Scott master cylinder) loses significant holding ability as tire radius increases.
~
Tire circumference has little to do with brake effectiveness. Weight being the same, brake effectiveness only relates to tire footprint and the distance the tire/surface contact is from the braking device..... and the brakes ability (hydraulic pressure) to stop the wheel from turning.
A single puck Cleveland brake (especially with stock Scott master cylinder) loses significant holding ability as tire radius increases.
~
FdxLou
BENEFACTOR
Canton, GA
On my Smith Cub with 212HP (modified 180) It was marginal with single piston Grove Brakes ( model 61-1 / 168,934 ft-lbs) on 26" Goodyears.
Going to 31" ABW was like having a brake failure on every flight.
I solved that by going to Grove double piston brakes ( model 66-154 / 280,188 ft-lbs). I have 1.50" axles. The improvement was impressive as a few of my landings at NH proved! I am now more confident landing on short( under 400ft) strips.
Looking back I should have gone to double piston Groves( better than Cleveland DP's imho...) when I switched from 850's to 26" GY.
Robbie Grove is a Super Cub owner and was great to work with. Call him if you have questions. www.groveaircraft.com.
Lou
Going to 31" ABW was like having a brake failure on every flight.
I solved that by going to Grove double piston brakes ( model 66-154 / 280,188 ft-lbs). I have 1.50" axles. The improvement was impressive as a few of my landings at NH proved! I am now more confident landing on short( under 400ft) strips.
Looking back I should have gone to double piston Groves( better than Cleveland DP's imho...) when I switched from 850's to 26" GY.
Robbie Grove is a Super Cub owner and was great to work with. Call him if you have questions. www.groveaircraft.com.
Lou
Clyde and Susan
BENEFACTOR
It is the tire/wheel radius that matters. The larger the radius the more leverage inertia and thrust has to overcome the brakes. So, the larger the wheel the more breaking power is needed when all else remains the same. None of this means that single puck brakes won't hold the airplane still with large tires. I have no idea how much single puck Clevelands can hold. ...Clyde Davis
bob turner
Registered User
Isn't the energy dissipation the same? You have to slow an object moving at 40 mph to zero - weighing in at, say, 1800 lbs, in 200 feet. Why would that take more energy dissipation with bigger tires?
That said, the big Grove/Cleveland ought to stop either one a lot quicker.
That said, the big Grove/Cleveland ought to stop either one a lot quicker.
Clyde and Susan
BENEFACTOR
Jack up the plane and remove the wheel. Replace the wheel with a plate attached to the break disk just like the wheel. Now attach a 6-inch pipe to the wheel replacement and have someone hold the brake. Try to make the brake disk rotate. Now add a longer length of pipe and try again. With a longer length of pipe (a longer lever) if it is long enough you will be able to over power the ability of the brake pads to hold the brake disk still. It is the same thing as when you are trying to loosen a nut with a short wrench and you have to use a longer wrench to make it move.
When you said that stopping the airplane would be dissipating the same amount of energy regardless of the tire size is correct. However, with a larger tire that same amount of energy is dissipated over a longer period of time and the airplane will travel further before stopping. With the longer lever arm provided by the larger tire, it will take more brake pad pressure to hold the airplane when an appreciable amount of power is advanced or to stop it from a given speed and weight. The longer lever arm provides more torque. The radius of the wheel is equal to the lever arm. ...Clyde
When you said that stopping the airplane would be dissipating the same amount of energy regardless of the tire size is correct. However, with a larger tire that same amount of energy is dissipated over a longer period of time and the airplane will travel further before stopping. With the longer lever arm provided by the larger tire, it will take more brake pad pressure to hold the airplane when an appreciable amount of power is advanced or to stop it from a given speed and weight. The longer lever arm provides more torque. The radius of the wheel is equal to the lever arm. ...Clyde
It takes more braking energy with bigger tires because to slow down you first must dissipate the energy stored in the tire and bigger=more energy (and the weight is a big deal here as is the increased moment arm of the increased diameter) before the tires go slower than the ground which has the desired effect of slowing the aircraft.bob turner said:
bob turner
Registered User
Not sure I buy that. The big heavy tire is at zero RPM before touchdown. Any energy imparted to the spinning wheel must have come at the expense of total aircraft energy.
Re: a post several up from here - Kinetic energy is simplified to force x distance. Seems to me that an aircraft at a given weight and speed is going to have a very specific amount of energy, and if you assume identical stopping distances, the energy dissipated must be the same. If the single puck brake is capable of doing it with the little tire, the theory says it ought to be able to do it with the big one. You may have to push harder, since the disc is moving slower, but that is a different calculation.
Re: a post several up from here - Kinetic energy is simplified to force x distance. Seems to me that an aircraft at a given weight and speed is going to have a very specific amount of energy, and if you assume identical stopping distances, the energy dissipated must be the same. If the single puck brake is capable of doing it with the little tire, the theory says it ought to be able to do it with the big one. You may have to push harder, since the disc is moving slower, but that is a different calculation.
jcrowles
Registered User
bemidji,mn.56601
Thirty years ago , I owned a number of Stinsons with the factory installed worthless Goodyear brakes. If you really wanted to get your heart thumping fast, You took the 108-3 out flying and then shot a 90* crosswind landing in a 15 MPH. wind. The brake got hard and the brakes got hot and round and around you went. I got pi**ed and installed a set of double puck clevelands that weren't approved at that time! What a differance that made !!! Same airplane, 90* x-wind, 25 MPH. and no trouble with directional control. I won't own a Stinson without Clevelands on them now!! My 2 cents worth !
aviationinfo
Registered User
Southwest WA
Are you going with single puck brakes because of weight? How much weight savings are we talking here?
FWIW, I once flew a J-5 with double puck Clevelands (Scott master cylinders and NOT boosted). My plane (PA-12) is 150+ lbs. heavier with the same brake system but the addition of boosters----the braking effectiveness is a LOT better.
Not sure what that tells you about single vs double pucks!
Andrew
FWIW, I once flew a J-5 with double puck Clevelands (Scott master cylinders and NOT boosted). My plane (PA-12) is 150+ lbs. heavier with the same brake system but the addition of boosters----the braking effectiveness is a LOT better.
Not sure what that tells you about single vs double pucks!
Andrew
Clyde and Susan
BENEFACTOR
Bob Turner, you're right. Big tire or small tire the same amount of energy must be dissipated. But, because of the larger tire radius you will either require more brake pad pressure (or more friction surface) to stop in the same distance or with the same brake pad surface and pressure more distance to dissipate that energy and stop. Perhaps the only way to convince you would be for you to perform an empirical test. ...Clyde
Bob Breeden
Registered User
Alaska
I wrote in above - I am pleased with single pucks on the 180 with a low radius/circumference tire, often I used ALL OF the braking power of double puck Clevelands, with brake boosters, on my Cub with 26's. So I was a bit disappointed with the braking when I switched to 31's. I'd like to improve on it - perhaps higher PSI brake boosters????
To Clarify - single pucks would be OK, but double pucks are superior for greater radius/circumference tires.
Bob Breeden
www.AlaskaAirpark.com
To Clarify - single pucks would be OK, but double pucks are superior for greater radius/circumference tires.
Bob Breeden
www.AlaskaAirpark.com
Longwinglover
Registered User
Charlotte, NC
Bob Breeden said:
Smaller tires???? :lol:
8) John Scott
Bob Breeden
Registered User
Alaska
Gordon Misch
MEMBER
Toledo, Wa (KTDO)
Oh boy! A topic I feel competent in (there aren't many), it's FRIDAY, and I just finished a bunch of very distasteful bureaucratic paperwork, so am in a good mood!
Re the posts about the energy: It may or may not be true that the bigger tire will absorb more energy from the airplane upon touchdown - that is, how much of the linear kinetic energy of the airplane is converted to rotational kinetic energy of the tire/wheel. It depends on how the mass of the tire varies with its loaded radius, and on the mass distribution. Both are unknowns here.
The rotational kinetic energy for different size wheels varies as 1/2 times the moment of inertia of the wheel times its angular velocity (related to RPM) squared. But for a given forward speed its angular velocity is inversely proportional to its loaded radius. Grossly simplifying, the tire/wheel assemblies will have a mass ratio roughly proportional to their diameter ratios, crudely assuming most of the mass is in the tread face of the tire. Now, moment of inertia is mr^2, so the ratio of moments of inertia will be as the ratio of their radii cubed (mass varies as r and I as r^2). Now, the angular velocity is proportional to v/r where v is the airplane's velocity and r is the loaded tire radius. so now we've got kinetic energy of the tire assembly roughly proportional to tire radius squared for a given linear velocity of the airplane.
HOWEVER that only has to do with initially spinning up the tire (and explains why the plane will lurch a little bit more on touchdown with the bigger tires). BUT the plane isn't stopped yet - the wheel needs to stop rotating when the plane stops. So the rotational kinetic energy of the wheel, as well as what's left of the linear kinetic energy of the plane, must also be converted to heat by the brakes.
In fact, it gets worse. A little bit of energy is dissipated by the wheel bearings. The larger wheel spins slower than a smaller wheel. Therefore the frictional energy loss in the wheel bearings is less for the large tire. This is a tiny difference, but it's there. So more of the total energy of the plane is dissipated in the brakes (as compared to other mechanisms) with the larger tires.
Bottom line - - Bob's right. No net savings in braking ENERGY for larger tires. But the braking energy only matters in terms of the heat dissipation capability of the brake system, not the braking force that must be applied to the pads.
As to adequacy of braking force, other posters who have said that the distance from ground to axle centerline is what matters, they're correct. The frictional force on the brake disk required for a given rate of deceleration is linearly proportional to that loaded radius.
Like I said, I'm more in a good mood right now than an intensely technical mood, so somebody may point out error(s) in the above. Fun stuff.
Re the posts about the energy: It may or may not be true that the bigger tire will absorb more energy from the airplane upon touchdown - that is, how much of the linear kinetic energy of the airplane is converted to rotational kinetic energy of the tire/wheel. It depends on how the mass of the tire varies with its loaded radius, and on the mass distribution. Both are unknowns here.
The rotational kinetic energy for different size wheels varies as 1/2 times the moment of inertia of the wheel times its angular velocity (related to RPM) squared. But for a given forward speed its angular velocity is inversely proportional to its loaded radius. Grossly simplifying, the tire/wheel assemblies will have a mass ratio roughly proportional to their diameter ratios, crudely assuming most of the mass is in the tread face of the tire. Now, moment of inertia is mr^2, so the ratio of moments of inertia will be as the ratio of their radii cubed (mass varies as r and I as r^2). Now, the angular velocity is proportional to v/r where v is the airplane's velocity and r is the loaded tire radius. so now we've got kinetic energy of the tire assembly roughly proportional to tire radius squared for a given linear velocity of the airplane.
HOWEVER that only has to do with initially spinning up the tire (and explains why the plane will lurch a little bit more on touchdown with the bigger tires). BUT the plane isn't stopped yet - the wheel needs to stop rotating when the plane stops. So the rotational kinetic energy of the wheel, as well as what's left of the linear kinetic energy of the plane, must also be converted to heat by the brakes.
In fact, it gets worse. A little bit of energy is dissipated by the wheel bearings. The larger wheel spins slower than a smaller wheel. Therefore the frictional energy loss in the wheel bearings is less for the large tire. This is a tiny difference, but it's there. So more of the total energy of the plane is dissipated in the brakes (as compared to other mechanisms) with the larger tires.
Bottom line - - Bob's right. No net savings in braking ENERGY for larger tires. But the braking energy only matters in terms of the heat dissipation capability of the brake system, not the braking force that must be applied to the pads.
As to adequacy of braking force, other posters who have said that the distance from ground to axle centerline is what matters, they're correct. The frictional force on the brake disk required for a given rate of deceleration is linearly proportional to that loaded radius.
Like I said, I'm more in a good mood right now than an intensely technical mood, so somebody may point out error(s) in the above. Fun stuff.
Clyde and Susan
BENEFACTOR
12 Geezer said:
Now that's what I was trying to say! ...Clyde
Gordon Misch
MEMBER
Toledo, Wa (KTDO)
Heheh, Clyde when I saw your name on that last post, before I read it, I thought you might've already found some big-ol dumb mistake in what I'd written! WHEW!!
Carey Gray
Registered User
Thun Field, Washington
delete
redrooster
Registered User
western Washington
FWIW, I've got double-pucks on my 170. I have run both Cleveland & Rapco brake pads in the past. p/n 66-105. Had similar results with braking action (good), wear (OK) and lining dust (lots). Last time around, I bought a set of the "Barry Jay" brake pads that Aircraft Spruce sells: "not faa/pma,experimental use only". Brakes work fine, they're wearing like iron, and very little to no dust around the brake area.
Current Spruce pricing for 8 pads:
Clevelands $82.40
Rapco $44.95
Barry Jay $18.95
Seems like a win/win to me.
Eric
Current Spruce pricing for 8 pads:
Clevelands $82.40
Rapco $44.95
Barry Jay $18.95
Seems like a win/win to me.
Eric
Longwinglover
Registered User
Charlotte, NC
[quote="NimpoCub] I learned early on, NOT to ask my Dad "What's the name of that star?".[/quote]
Nimpo,
I know what you mean! My mom knew all the movie stars names too! :lol:
John Scott
Nimpo,
I know what you mean! My mom knew all the movie stars names too! :lol:
John Scott
Ruidoso Ron
Registered User
Alto, NM
A BELLAND said:
Just be careful on that first landing. Braking will be significantly better than with the Goodyears. I shipped my new Clevelands to my brother, who installed them on my SC for me. On his first landing he managed to get it on its nose in soft dirt, and minimal damage was done. He said that he didn't know that an airplane brake could work that well.
bob turner
Registered User
Soft dirt can do that with no brakes.
There for a while, Barry Jay pads were eight for $12. Then Aircraft Spruce bought them out. Got me a lifetime supply first.
There for a while, Barry Jay pads were eight for $12. Then Aircraft Spruce bought them out. Got me a lifetime supply first.
