@@robhaylock7742there are quite a number of statements you make that are wrong! badly wrong The profile drag is also proportional to ½rho×V²×A, not at all a separate drag as falsely shown on your graph, it is part of the parasite drag. The induced drag does not level off. The parasite drag follows a parabola, at zero speed all drag is exactly zero, including your depicted "profile drag" , also of course the lift. L/D max is where the slope of the parasite drag is equal and opposite to the slope of the induced drag. Now, 1 exponential means Y(x) = e^x 2 quadratic means Y(x) = x² 3 cubic means Y(x) = x³ Parasite Drag = Sum of all parasite components, which are proportional to the q = ½rho V² A QUADRATIC function of true airspeed V Summarized Dp = q × Cdp × A A is a convenient reference area like the wing projected area Cdp depends on the direction of the flow and on the Reynolds number Re Re = V × c / nu nu = viscosity / density Called dynamic viscosity the viscosity depends on the temperature. c is a reference length mostly the mean chord or the length of the fuselage Typical value is tenths of millions See Horner's Fluid dynamic drag, a book that most aerodynamicists hàve and treasure. There you find all in correct form On a out of ground effect hovering helicopter the downwash accelerates to twice it's speed of the flow passing through the rotor plane. The inflow comes from the entire sphere around the rotor, except the narrow jet below. Half of the flow volume comes from the lower half and makes a sharp turn just outside the tip ring , the upper half of the flow passes through the plane of the rotor in the inner 70.7% of the rotor radius. A true flow field that your description misses considerably. If you do the analysis correctly that is what you would discover The correct induced drag and lift equations have been developed by McCormic and are described in terms of the final wake angle beta: Lift = q×π/4 ×b²×2×sin(beta)×(1+ cos(beta)²) Dind = q×π/4 ×b²×2×sin(beta)² ×cos(beta) b is span the parasite drag is q×CDA, the flat plate area times q, called dynamic pressure or incorrectly stagnation pressure these exact results differ at low indicated speeds considerably from your sketches You know the difference from true airspeed? At 12000 feet pressure altitude the true airspeed is 1.200 times the calibrated airspeed, when the outside temperature is -9C Remark: if you plot the 2 curves as a function of q, the dynamic pressure q = ½rho ×V² The parasite drag is a constant slope linear line The induced drag is a hyperbola and what is important: independent of the density, altitude, corresponding to calibrated airspeed, approximately indicated airspeed. Confused? Get a pilot certificate.
Arturo, there is PLENTY of drag in the hover, mainly induced drag. The helo needs more power in the hover because of this, and it can be reduced by Ground Effect changing the inflow angle. The power required decreases as translational lift is achieved, and the induced flow is reduced. There is also an element of parasite drag in the hover, with the downwards airflow over the fuselage causing some drag.
@@robhaylock7742 You are right about the power in hover But you are confusing power with longitudinal drag, drag of the blades produces a torque on the shaft. Multiplying this torque with rpm × 2π/60 gives you the required POWER. P = L^1.5 /D ×√(2/π rho) + 3/4 L× Tipspeed × CD/Cl In hover, no contribution of the fuselage since it is not moving. As you transition into forward flight , the powered lift diminishes as the translational lift increases, the rotor acting like the lift on an autogyro , the power that the rotor requires is reduced to a minimum, the best endurance speed, then the fuselage drag and the blade differential drag increase, while the induced drag due to translational lift decreases this drag is a force opposite to the direction of flight. induced drag applies only to longitudinal motion , a force, not a moment. the forward tilted rotor lift produces the thrust to over come the 3 drag forces involved. The power required to create the downwash in hover is not to overcome a longitudinal force but a moment due to blade drag and to accelerate the rotor down wash . You are confusing torque with drag. read Barnes McCormick, Aerodynamics of VTOL flight (Helicopter designer here.)
@@arturoeugster7228 In the hover, as the induced flow increases to produce more downwash (and get off the ground), the relative airflow comes onto the rotor blade from a higher angle compared to the plane of rotation - the lift component is tilted further back, and the drag component is pointing downwards as well as backwards. Basically, adding to the weight. This adds to the power required to hover. Lift that lever, see the torque meter go up. Getting into ground effect reduces the induced flow, brings the RAF in a bit lower, and the drag component acting downwards is reduced, meaning less power needed to hover. "The power required to create the downwash in hover is not to overcome a longitudinal force but a moment due to blade drag and to accelerate the rotor down wash ." What a confused statement. All downwash is due to induced flow. The air gotta come from above to get pushed downwards and create the lifting force. In the OGE hover, a column of air from above is induced into downwards motion increasing the IF and requiring a higher pitch angle to get the desired AoA. Get into forward flight, and the blade sees clean air which hasn't had a chance to start moving down much, so less IF at the front half of the disc. The back of the disc, where the blade is flapped up, gets more induced flow simply because it is up there, resulting in inflow roll. "You are confusing torque with drag." And how does the helicopter get the power to overcome the various drag bits? From torque. The helicopter in flight has one big arrow, at 90 degrees to the tip path plane, pointing forward and upward, opposed by one big arrow in the opposite direction pointing rearward and downward. Split these into components in the horizontal plane and vertical plane, you have Lift, opposed by Weight straight down, and Thrust forward, opposed by Drag straight back. Lift + Thrust equals Weight + Drag. Slow down and the arrows come back towards the vertical, until you are in the hover, straight up and down. You can say that they still show L+T=W+D, with the extra thrust needed to overcome the vertical component of drag, not just parasite, but the downward component of drag on each blade, as described above.. And the fuselage, in the hover, does see a component of parasite drag from the airflow coming down onto it. Read about the problems with the wings fitted to the French contender for battlefield helicopters. Bad in the hover, great in forward flight.
The power curve on a helicopter moves out almost constant, then drops gently to the max endurance speed and then rises cubically to high values
4:30 parasite drag increases quadraticaly, not exponentially
~ V². not. .e^V
Not sure what you mean by this - V squared is an exponential. Quadratic would mean drag increases at V to the fourth, which is not correct.
@@robhaylock7742there are quite a number of statements you make that are wrong! badly wrong
The profile drag is also proportional to ½rho×V²×A, not at all a separate drag as falsely shown on your graph, it is part of the parasite drag. The induced drag does not level off.
The parasite drag follows a parabola, at zero speed all drag is exactly zero, including your depicted "profile drag" , also of course the lift.
L/D max is where the slope of the parasite drag is equal and opposite to the slope of the induced drag.
Now,
1 exponential means
Y(x) = e^x
2 quadratic means
Y(x) = x²
3 cubic means
Y(x) = x³
Parasite Drag = Sum of all parasite components, which are proportional to the q = ½rho V²
A QUADRATIC function of true airspeed V
Summarized
Dp = q × Cdp × A
A is a convenient reference area like the wing projected area
Cdp depends on the direction of the flow and on the Reynolds number Re
Re = V × c / nu
nu = viscosity / density
Called dynamic viscosity
the viscosity depends on the temperature.
c is a reference length mostly the mean chord or the length of the fuselage
Typical value is tenths of millions
See Horner's Fluid dynamic drag, a book that most aerodynamicists hàve and treasure.
There you find all in correct form
On a out of ground effect hovering helicopter the downwash accelerates to twice it's speed of the flow passing through the rotor plane.
The inflow comes from the entire sphere around the rotor, except the narrow jet below. Half of the flow volume comes from the lower half and makes a sharp turn just outside the tip ring , the upper half of the flow passes through the plane of the rotor in the inner 70.7% of the rotor radius.
A true flow field that your description misses considerably.
If you do the analysis correctly that is what you would discover
The correct induced drag and lift equations have been developed by McCormic and are described in terms of the final wake angle beta: Lift = q×π/4 ×b²×2×sin(beta)×(1+ cos(beta)²)
Dind = q×π/4 ×b²×2×sin(beta)² ×cos(beta)
b is span
the parasite drag is q×CDA, the flat plate area times q, called dynamic pressure or incorrectly stagnation pressure
these exact results differ at low indicated speeds considerably from your sketches
You know the difference from true airspeed?
At 12000 feet pressure altitude the true airspeed is 1.200 times the calibrated airspeed, when the outside temperature is -9C
Remark:
if you plot the 2 curves as a function of q, the dynamic pressure
q = ½rho ×V²
The parasite drag is a constant slope linear line
The induced drag is a hyperbola and what is important: independent of the density, altitude, corresponding to calibrated airspeed, approximately indicated airspeed.
Confused? Get a pilot certificate.
The drag curve shown applies to airplanes, not to powered helicopters!
No drag in hover, but plenty of POWER
Arturo, there is PLENTY of drag in the hover, mainly induced drag. The helo needs more power in the hover because of this, and it can be reduced by Ground Effect changing the inflow angle. The power required decreases as translational lift is achieved, and the induced flow is reduced. There is also an element of parasite drag in the hover, with the downwards airflow over the fuselage causing some drag.
@@robhaylock7742 You are right about the power in hover
But you are confusing power with longitudinal drag, drag of the blades produces a torque on the shaft. Multiplying this torque with rpm × 2π/60 gives you the required POWER.
P = L^1.5 /D ×√(2/π rho) + 3/4 L× Tipspeed × CD/Cl
In hover, no contribution of the fuselage since it is not moving.
As you transition into forward flight , the powered lift diminishes as the translational lift increases, the rotor acting like the lift on an autogyro , the power that the rotor requires is reduced to a minimum, the best endurance speed, then the fuselage drag and the blade differential drag increase, while the induced drag due to translational lift decreases
this drag is a force opposite to the direction of flight.
induced drag applies only to longitudinal motion , a force, not a moment. the forward tilted rotor lift produces the thrust to over come the 3 drag forces involved. The power required to create the downwash in hover is not to overcome a longitudinal force but a moment due to blade drag and to accelerate the rotor down wash .
You are confusing torque with drag.
read Barnes McCormick, Aerodynamics of VTOL flight
(Helicopter designer here.)
@@arturoeugster7228 In the hover, as the induced flow increases to produce more downwash (and get off the ground), the relative airflow comes onto the rotor blade from a higher angle compared to the plane of rotation - the lift component is tilted further back, and the drag component is pointing downwards as well as backwards. Basically, adding to the weight. This adds to the power required to hover. Lift that lever, see the torque meter go up.
Getting into ground effect reduces the induced flow, brings the RAF in a bit lower, and the drag component acting downwards is reduced, meaning less power needed to hover.
"The power required to create the downwash in hover is not to overcome a longitudinal force but a moment due to blade drag and to accelerate the rotor down wash ."
What a confused statement. All downwash is due to induced flow. The air gotta come from above to get pushed downwards and create the lifting force. In the OGE hover, a column of air from above is induced into downwards motion increasing the IF and requiring a higher pitch angle to get the desired AoA. Get into forward flight, and the blade sees clean air which hasn't had a chance to start moving down much, so less IF at the front half of the disc.
The back of the disc, where the blade is flapped up, gets more induced flow simply because it is up there, resulting in inflow roll.
"You are confusing torque with drag." And how does the helicopter get the power to overcome the various drag bits? From torque.
The helicopter in flight has one big arrow, at 90 degrees to the tip path plane, pointing forward and upward, opposed by one big arrow in the opposite direction pointing rearward and downward. Split these into components in the horizontal plane and vertical plane, you have Lift, opposed by Weight straight down, and Thrust forward, opposed by Drag straight back. Lift + Thrust equals Weight + Drag. Slow down and the arrows come back towards the vertical, until you are in the hover, straight up and down. You can say that they still show L+T=W+D, with the extra thrust needed to overcome the vertical component of drag, not just parasite, but the downward component of drag on each blade, as described above..
And the fuselage, in the hover, does see a component of parasite drag from the airflow coming down onto it. Read about the problems with the wings fitted to the French contender for battlefield helicopters. Bad in the hover, great in forward flight.