probably a silly question, but I am interested !. Got some details on the prop fitted to a Griffon 66. the engine ticks over at 500 rpm, the prop turning at a lazy 255rpm, and when opened up to max engine revs of 2750 the prop is wazzing round at 1402.5 rpm.
Now the prop diameter is 10′ 5″ and the tip speed at tick over is 95mph and 522mph at full chat.
My question, in a few parts is…finally … what does a single prop blade weigh stationary, whats the centrifugal force (in Lbs) at tickover and whats the higher force when the loud lever is firewalled…in other words how much does the prop weigh when its being used ?.
Must be a hell of a stress on the hub !
Anybody any good at maths ?
By: Rocketeer - 17th May 2005 at 23:38
OK, a crude stab without the calculus ( mass varies along the blade length as does speed )
centripetal force = mass x acceleration
acceleration = V squared / radius
so centripetal force = mass x V squared / radiuscrude assumptions:
centre of mass is 25% blade length from the hub centre and mass is 25kgthis comes out with a radius of 0.4 m, a speed ( V ) of about 6 m/s and a mass of 25kg
This works out at 2156 Newtons, 220 kg or 484lbs centripetal force.
Anyone agree/disagree ?
It is 23:30, I am tired, I cannot believe I am having a go at this!!!
Using your assumptions…..
Centripetal F = mr(omegaSqd), where v=romega
substituting gives F=mr(v2/r2), so F=mv2/r
using metric, F=25×36/0.4=2250N…..225kg…..so yep youre right…..with your assumptions!
however, 2500rpm is v=romega=0.4x2500x2pi=6282rad/min=104.7rad/sec…..cant be bothered!!!
By: italian harvard - 17th May 2005 at 21:39
You lost me at “OK”…
eheheheh 😀
Alex
By: geedee - 17th May 2005 at 21:35
OK, a crude stab without the calculus ( mass varies along the blade length as does speed )
centripetal force = mass x acceleration
acceleration = V squared / radius
so centripetal force = mass x V squared / radiuscrude assumptions:
centre of mass is 25% blade length from the hub centre and mass is 25kgthis comes out with a radius of 0.4 m, a speed ( V ) of about 6 m/s and a mass of 25kg
This works out at 2156 Newtons, 220 kg or 484lbs centripetal force.
Anyone agree/disagree ?
So times that by six blades….blimey, thats a lot of force trying to pull that hub apart,
hmmmmm thanks for that mate.
By: Slipstream - 17th May 2005 at 21:12
OK, a crude stab without the calculus ( mass varies along the blade length as does speed )
centripetal force = mass x acceleration
acceleration = V squared / radius
so centripetal force = mass x V squared / radius
crude assumptions:
centre of mass is 25% blade length from the hub centre and mass is 25kg
this comes out with a radius of 0.4 m, a speed ( V ) of about 6 m/s and a mass of 25kg
This works out at 2156 Newtons, 220 kg or 484lbs centripetal force.
Anyone agree/disagree ?
By: italian harvard - 17th May 2005 at 20:34
we tried with colored crayons, with a color for each subject, but the lack of thumbs makes the thing difficult.. after a while u get used to read pee wet newspapers..
Alex
By: Melvyn Hiscock - 17th May 2005 at 20:30
my dogs doesnt just bring me the newspaper, he even circles the most interesting articles..
Alex
What with?
By: geedee - 17th May 2005 at 20:29
Ah, you should have got a cleverer dog.
The spat shaped hats don’t help.
All I could think of when that was on my head was how much water I had to cross later in the day!
The answer to your question is . . . . . . . . lots.
well there were 2 dogs telling me how to repair the car one day when it broke down. 1 dog was white, the other black and they both kept telling me different things. this freaked me out so legged it to the village pub and recounted my tale. the barman said to ignore the white dog cos he knew nothing about cars, so I went back and got the black …read smarter…dog. Methinks I wus diddled !
Course, the next question is …how far would the blade travel if it came ‘unstuck’ !
By: italian harvard - 17th May 2005 at 20:25
my dogs doesnt just bring me the newspaper, he even circles the most interesting articles..
Alex
By: Melvyn Hiscock - 17th May 2005 at 20:23
Got one…. take it walks….asked it and it doesnt know, hence the question on the forum where people know these answers….says he reaching for a red spat shaped hat to concentrate his thoughts
Ah, you should have got a cleverer dog.
The spat shaped hats don’t help.
All I could think of when that was on my head was how much water I had to cross later in the day!
The answer to your question is . . . . . . . . lots.
By: italian harvard - 17th May 2005 at 20:12
u need an engineer for dis kinda questions.. I guess a Griffon Spit blade might weight some 50kg? If u want to be sure find a blueprint of the prop, get the size of the thing and apply the specific weight of aluminium 😉
Alex
By: geedee - 17th May 2005 at 19:44
Got one…. take it walks….asked it and it doesnt know, hence the question on the forum where people know these answers….says he reaching for a red spat shaped hat to concentrate his thoughts
By: Melvyn Hiscock - 17th May 2005 at 19:37
probably a silly question, but I am interested !. Got some details on the prop fitted to a Griffon 66. the engine ticks over at 500 rpm, the prop turning at a lazy 255rpm, and when opened up to max engine revs of 2750 the prop is wazzing round at 1402.5 rpm.
Now the prop diameter is 10′ 5″ and the tip speed at tick over is 95mph and 522mph at full chat.
My question, in a few parts is…finally … what does a single prop blade weigh stationary, whats the centrifugal force (in Lbs) at tickover and whats the higher force when the loud lever is firewalled…in other words how much does the prop weigh when its being used ?.Must be a hell of a stress on the hub !
Anybody any good at maths ?
You have too much time on your hands.
Buy a dog
Take it for walks.
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May 17, 2005 at 7:20 pm