Perry Mick wrote:
snips.....
Could it be that the Ross redrive is somehow harder on these
bearings than a prop is?
I'm thinking of reassembling it and measuring the end play so
we have some numbers. Then replacing the bearings anyway since
the set is only $50, then send the old one's to you to inspect.
Tracy Crook wrote:
Yes, there is an enormous diference between your direct drive and the forces
imposed by the helical gears in the Ross drive. I don't have the exact
formula in front of me right now but the force from the gear is something
like Max engine torque (in foot pounds) divided by input gear radius (in
feet) times cosign of the gear helix angle (22 degrees). This calculates
out to about 800 pounds. At 200 mph and assuming 200 HP, the prop thrust is
far less than 300 pounds. Running direct drive, the horsepower (and thrust)
would be even less.
Tracy Crook
rws@altavista.net
Paul wrote:
Applying our HP formula... if the prop efficiency is 80% the thrust at 200 MPH is
is 160 HP net out of the prop times 550 divided by FPS or 293 FPS or 300 pounds.
Very good guess Tracy.
The worst case is full throttle climb at 100 MPH (146 FPS) or below where 160 net HP
out of the prop translates into at least 600 pounds of thrust assuming the engine
is running at max HP RPM.
Paul wrote:
Here are a few other things we can figure out from the basic HP formula.
We can estimate the thrust from the top speed data and the climb data.
If the airplane goes 200 MPH on 160 HP net out of the prop the
drag on the airplane is 300 pounds. Since drag = flat plate area
times speed squared times .0026 the flat plate area must be;
flat plate = 300/(MPH^2 * .0026) = 300/104 = 2.9 square feet.
Induced drag due to lift is low at this speed and can be ignored.
The .0026 is the sea level air density constant when MPH is used.
At 100 MPH the drag is then 75 pounds. The net HP out of the prop to over come drag
is
= (146 * 75)/550 or 20 HP Actually I would say it is a bit more than this
as we are getting down at a speed where induced drag due to lift is becoming
a major factor. Lets say for argument the airplane needs 40 HP to fly straight
and level at 100 MPH. The minimum drag speed for an RV is about 135 MPH as
I recall.
Now the airplane is climbing at 1500 feet per minute at 100 MPH and
the gross weight is 1600 pounds. So that is 25 feet per second straight up.
This is like the horse lifting a 1600 pound weight with a pulley arrangement.
Actually it is going to take a few horses to lift this amount of weight at 25 FPS.
The HP requirement is (1600 lb * 25 FPS) / 550 or(40000 lb ft sec)/550
or about 73 HP.
Total HP needed to climb out at 100 MPH and 1500 FPM is therefor about 113 HP.
113 HP = (thrust * 146 FPS)/550... doing the algebra..
113 HP * 550 = (thrust * 146 FPS)... doing the algebra..
Thrust = (113 HP * 550)/146 FPS or 426 pounds of thrust roughly.
Perry Mick wrote:
I'm glad you guys know how to calculate this stuff, especially the
forces imposed by the helical gears, I don't have a clue there. Looks like
Tracy hit on the source of the problem with the Ross drive and the stock
thrust bearings.
With my weight of 1400 lbs. and climb rate of 1000 fpm, my thrust is
even less than Paul's numbers:
climb rate = 17 fps
HP requirement = 1400 lb * 17 fps / 550 = 43 HP
adding the HP for drag = 40 + 43 HP = 83 HP
Thrust = 83 * 550 / 146 = 312 lbs.
This is pretty close to my prop analysis program which shows about 350
lbs. of thrust static. But the plot shows 275 lbs of thrust at 90 knots,
which is prop only I think, so the extra 37 lbs. of thrust may be duct thrust.
This also says the stock thrust bearings may be a problem in direct
drive with higher HP and thrust levels than I'm achieving. But according to
Charles Harbert the engine can be modified for better thrust bearings so
there is always that option.
--
Perry J. Mick
Mick Duckt N7XR
http://www.ductedfan.com
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