HumanPoweredDesigner
Well-Known Member
I hope equations 1 or 2 for human powered helicopter lift are wrong, and that the diameter does not have to be over 150 feet.
Let's assume 100% efficiency, and that a hovering gets its lift by producing a slip stream cylinder of air of constant velocity with respect to radius.
Thrust is proportional to momentum transfer per second: mass flow rate x velocity,
and Power is proportional to kinetic energy transfer per second which is thrust x velocity.
Since Power has more velocity terms multiplied together than thrust does, you get the most thrust at the least velocity when velocity of the whole cylinder is equal across the disk.
While 2+2 is the same total thrust in the disc as 3+1,
2*2+2*2 is less power than 3*3+1*1, and power is actually the third power and thrust the second power, so it would be 2*2*2+2*2*2 is less power than 3*3*3+1*1*1, a greater than 50% increase.
Some short hand:
A coefficient after a term means an exponent, whereas before the term it means a coefficient.
kg = kilogram
M = gross mass of helicopter in kilograms
m = meter
m2 = square meter, m3 = cubic meter
s = seconds
S = surface area of ideal disk cross section in m2 = pi*r2
P = Power in watts, kg*m2/s3
p= density air at sea level and 60 degress = 1.2 kg/m3
v = velocity of air exiting disk in m/s
g = gravity = 9.8 m/s2
L = Lift in neutons, kg*m/s2
1. P = L*v = M*g*v
2. L = M*g = S*p*v*v (the units match)
If we know P and M, we can solve for v and S, since P and M become coefficients in two equations with two unknowns.
From equation 1., 3. v = P/(M*g)
If we plug v from equation 3 into equation 2 to and solve for S, we get
4. (M*g)3/(P2*p) = S
For a 135 pound pilot and 80 pound helicopter, I'm getting a 26 meter radius needed if the pilot can generate 600 watts for 60 seconds. That is 2.6x the Yuri radius, and is over 150 feet in diameter. Deep ground effect and a 100 pound pilot may be the reason the yuri got off the ground for 20 seconds. 10 feet is still in ground effect, though.
Equation 4 also states that if you double the radius, the power needed is halfed.
If you double the mass, the power needed goes up by 2 to the 1.5 power.
But in order for the Power to be minimized, the whole disk must have uniform velocity. If the rpm is too low, you will have high velocity trailing the blade, followed by lower velocity soon after. With tip losses and hub voids, and low rpm, the thrust will not be uniform, and larger radius means it is harder to get high rpm.
Let's assume 100% efficiency, and that a hovering gets its lift by producing a slip stream cylinder of air of constant velocity with respect to radius.
Thrust is proportional to momentum transfer per second: mass flow rate x velocity,
and Power is proportional to kinetic energy transfer per second which is thrust x velocity.
Since Power has more velocity terms multiplied together than thrust does, you get the most thrust at the least velocity when velocity of the whole cylinder is equal across the disk.
While 2+2 is the same total thrust in the disc as 3+1,
2*2+2*2 is less power than 3*3+1*1, and power is actually the third power and thrust the second power, so it would be 2*2*2+2*2*2 is less power than 3*3*3+1*1*1, a greater than 50% increase.
Some short hand:
A coefficient after a term means an exponent, whereas before the term it means a coefficient.
kg = kilogram
M = gross mass of helicopter in kilograms
m = meter
m2 = square meter, m3 = cubic meter
s = seconds
S = surface area of ideal disk cross section in m2 = pi*r2
P = Power in watts, kg*m2/s3
p= density air at sea level and 60 degress = 1.2 kg/m3
v = velocity of air exiting disk in m/s
g = gravity = 9.8 m/s2
L = Lift in neutons, kg*m/s2
1. P = L*v = M*g*v
2. L = M*g = S*p*v*v (the units match)
If we know P and M, we can solve for v and S, since P and M become coefficients in two equations with two unknowns.
From equation 1., 3. v = P/(M*g)
If we plug v from equation 3 into equation 2 to and solve for S, we get
4. (M*g)3/(P2*p) = S
For a 135 pound pilot and 80 pound helicopter, I'm getting a 26 meter radius needed if the pilot can generate 600 watts for 60 seconds. That is 2.6x the Yuri radius, and is over 150 feet in diameter. Deep ground effect and a 100 pound pilot may be the reason the yuri got off the ground for 20 seconds. 10 feet is still in ground effect, though.
Equation 4 also states that if you double the radius, the power needed is halfed.
If you double the mass, the power needed goes up by 2 to the 1.5 power.
But in order for the Power to be minimized, the whole disk must have uniform velocity. If the rpm is too low, you will have high velocity trailing the blade, followed by lower velocity soon after. With tip losses and hub voids, and low rpm, the thrust will not be uniform, and larger radius means it is harder to get high rpm.
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