# Estimation of minimum power required

#### XXavier

##### Member
Chatting with a friend, he touched the issue of the minimum power required by a gyro to stay aloft. I didn't know an answer, but thinking about that question, I have found the following (tentative) answer. Any comments will be welcome...

I remember having read somewhere that an autogyro, in vertical autorotation, has a sink rate more or less like a parachute having the diameter of the rotor, and loaded with the same mass of the gyro.

I've found, in Wikipedia, the formula for the terminal velocity of a body of mass m, area s, falling with a drag coefficient Cd in air of density rho:
v = 1,41 · m^0,5 · g^0,5 · rho^–0,5 · s^–0,5 · Cd^–0,5

I get v = 9,3 m/s for an ELA gyro, with a mass of 400 kg and a rotor diameter of 8,5 m and assuming Cd = 1,3

That vertical autorotation speed sounds plausible... It's 1830 fpm... Now, for a descending parachute at a constant speed (or for a gyro in vertical autorotation), the drag (or the rotor lift) is equal in magnitude to the weight. The power associated with that moving force is m · g · v. For the ELA, that's 400 · 9,8 · 9,3 = 36456 W = 49,5 hp. It looks a bit high for minimum power required, since the efficiency of the prop may be around 0,6. That would mean a shaft power of 83 hp...

As stated above, any comments (or corrections) will be welcome...

#### C. Beaty

##### Gold Supporter
Here’s a plot I made many years ago for a Bensen type gyro. Airframe drag is an estimate based on an equivalent flat plate area. Drawbar power is the power required with a propeller efficiency of 100%.

Unfortunately, the math program I used for this calculation was specific to WXP.

#### C. Beaty

##### Gold Supporter
Calculations based on rate of vertical descent would be the power required to sustain a hovering helicopter that had a rotor efficiency of 100%.

#### XXavier

##### Member
Here’s a plot I made many years ago for a Bensen type gyro. Airframe drag is an estimate based on an equivalent flat plate area. Drawbar power is the power required with a propeller efficiency of 100%.

Unfortunately, the math program I used for this calculation was specific to WXP.
View attachment 1145276

I see that the minimum power required for staying aloft was 12-13 hp. I suppose it was calculated from the 'drawbar power'. Probably a gyro glider towed by a car with a dynamometer... That's quite low, even for a lighter gyro as the Bensen...

#### XXavier

##### Member
Calculations based on rate of vertical descent would be the power required to sustain a hovering helicopter that had a rotor efficiency of 100%.

Leaving apart the rotor efficiencies, the power required to sustain a hovering helicopter can't be different from the power exerted by the same rotor, of the same helicopter, while descending at a constant speed in autorotation, since lift and weigh are exactly balanced also in the latter case. It sounds surprising, but the fact is that if lift and weight weren't exactly balanced, the net difference in force will cause an acceleration, and we have zero acceleration here...

#### Gyro28866

##### David McCutchen
"since lift and weigh are exactly balanced also in the latter case."
.

If lift and weight are equal, that means it is in a steady state and is not increasing or decreasing the vertical plane of what we call altitude.
hmmm? Just a thought:
Is our thinking confusing us?
When in a Vertical Descent, the rotors rpm is in a steady state because of gravities constant. For the rotor to maintain autorotation it has to have the relative wind flow within a range of degrees inflow. So, if the relative wind inflow actually is in the vertical and our forward velocity actually is in the vertical also? Then, if you think about this - while in the vertical descent; is the rotors 6 oclock position actually the leading edge for the relative wind inflow? And are we just flying backwards

#### XXavier

##### Member
"since lift and weigh are exactly balanced also in the latter case."
.

If lift and weight are equal, that means it is in a steady state and is not increasing or decreasing the vertical plane of what we call altitude.
hmmm? Just a thought:
Is our thinking confusing us?
When in a Vertical Descent, the rotors rpm is in a steady state because of gravities constant. For the rotor to maintain autorotation it has to have the relative wind flow within a range of degrees inflow. So, if the relative wind inflow actually is in the vertical and our forward velocity actually is in the vertical also? Then, if you think about this - while in the vertical descent; is the rotors 6 oclock position actually the leading edge for the relative wind inflow? And are we just flying backwards

If lift and weight are equal, there's no acceleration. It's also the case of a parachute, where upwards-directed drag balances weight exactly, hence acceleration is zero, and the sink speed is constant...

#### C. Beaty

##### Gold Supporter
The drag of a rotor or wing has two components, profile drag and induced drag.

Profile drag is the drag that results from moving an airfoil through the air and increases as V². The primary source of profile drag of a rotor is the result of rotation.

Induced drag is the result of “inducing” a downward acceleration in the surrounding air mass to produce lift in the same manner that a shotgun produces recoil by accelerating shot down the barrel.

It decreases as 1/V².

#### kolibri282

##### Super Member
The 83 hp you get are way to high in my opinion. If you ask for the minimum power to stay aloft you should base your calculation on the minimum sink rate, which a gyro and a helicopter attain at a certain forward speed. This is due to the fact that in hover or vertical descent a lot of power is required to overcome the induced drag. Induced flow through the rotor becomes smaller with greater forward speed. This is the reason why early models of the S-58 could lift off with a much higher load when they used their landing gear to gather forward speed by rolling before lift off. With the same load they were not able to hover even in ground effect.

#### XXavier

##### Member
The 83 hp you get are way to high in my opinion. If you ask for the minimum power to stay aloft you should base your calculation on the minimum sink rate, which a gyro and a helicopter attain at a certain forward speed. This is due to the fact that in hover or vertical descent a lot of power is required to overcome the induced drag. Induced flow through the rotor becomes smaller with greater forward speed. This is the reason why early models of the S-58 could lift off with a much higher load when they used their landing gear to gather forward speed by rolling before lift off. With the same load they were not able to hover even in ground effect.

Thanks...

Yes, I have done that several times, with gyros and FWs, gliding in a calm day, with the engine idling, and slowing down little by little, until findng the IAS for least sink rate. That minimum sink rate, times the weight of the aircraft, gives the minimum power required in 'gravitational energy' terms. Correcting for prop efficiency, one can get a reasonable estimate of the minimum (shaft) power required.

But I was trying to find that minimum power required from (more or less) first principles. After reading your post, I think that perhaps a way would be to compute (in simplified, approximate terms) the theoretical lift and drag of the gyro for three or four airspeeds, then, with that data, to construct a curve of L/D as a function of airspeed, and finally, in some way, infer from the curve the 'certain forward speed' that you mention, and from it the minimum power required. (All that is just 'thinking aloud' right now...)

Of course, I know that all this has been solved, ages ago, by very competent people, and is somewhere 'in the books', but all this is anyway a good mental exercise for aging brains, and I'm already in my 70s...

#### C. Beaty

##### Gold Supporter
Scan from Bensen flight training manual:

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#### C. Beaty

##### Gold Supporter
With this chart, Bensen was explaining, for the low time pilot flying at some speed slower than the best rate of climb speed that to clear an obstacle, contrary to instinct, push the stick forward and dive to reach best rate of climb speed and then pull back on the stick to climb over the object.

#### kolibri282

##### Super Member
Quote: But I was trying to find that minimum power required from (more or less) first principles /Quote

This requires to write a computer program that uses the formulae in e.g. naca 716 (where lift and drag of a rotor for rotary wing aircraft is calculated from first principles) for the gyro and another set that describes the thrust generated by an engine to which a propeller of given geometry (diameter and advance ratio) is attached add quite some more stuff and then program what is called an iterative solution to the trim problem. Currently (to the best of my knowledge) there area three such open source programs available:

Jeans program written in JAVA, if you would like a version that runs in the eclipse IDE drop me a line with a private e-mail, I could send you a workspace for the Mars version of eclipse where I got the program up and running

b) Bruno Zilli:
Brunos program uses scilab, a free matlab/simulink clone, it runs out of the box in scilab.

c) kolibri282:
This last program uses octave, a free matlab clone. Note that the divshare link in c does not work anymore. If you are interested in program c again drop me an e-mail address

PS: the drawing shown here as my profile picture in the forum is one of many schematic drawings I made during the (very slow) process of getting to grips with the plethora of variables involved in rotary wing flight.....;-)
PPS: I seem to remember that Bruno's program just a couple pages but if you want to take a look at a) or c) take a deep breath before you open the zip file, these consist of several thousand lines of code....

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#### Vance

##### Gyroplane CFI
Another way of looking at it.

The Predator has a Lycoming IO-320 rated at 160 horsepower and according to the manual burns around .45 gallons per horsepower hour.

The Predator at an all up weight of 1,400 pounds burns around 6 gallons per hour at 45kts and a gallon of gas weighs abound six pounds.

The Predator is burning 36 pounds of fuel per hour divided by .45 and she requires around 80 horsepower to fly at 45kts indicated air speed (52 miles per hour).

Increase the speed to 65kts indicated air speed (75mph) and she burns closer to 7 gallons per hour so she is making around 93 horsepower.

Climb to 7,500 feet density altitude and at 90kts indicated air speed (104 mph) she burns around ten gallons per hour making 133 horsepower based on these simple calculations. I suspect it is a little less because she is using some of the fuel to keep her cool at wide open throttle.

This is around 104kts (120mph) true air speed.

#### XXavier

##### Member
Quote: But I was trying to find that minimum power required from (more or less) first principles /Quote

This requires to write a computer program that uses the formulae in e.g. naca 716 (where lift and drag of a rotor for rotary wing aircraft is calculated from first principles) for the gyro and another set that describes the thrust generated by an engine to which a propeller of given geometry (diameter and advance ratio) is attached add quite some more stuff and then program what is called an iterative solution to the trim problem. Currently (to the best of my knowledge) there area three such open source programs available:

Jeans program written in JAVA, if you would like a version that (...)

(...)
(...)
(...)

Lots of thanks, Juergen, for the extensive information that you supply. I use computers, of course, but I don't go beyond some simple programs in BASIC. Using MATLAB is beyond my abilities. I know because I tried...

If I'm working in this things, trying to get them 'from first principles', it's just for fun (and in order to exercise my aging brain...). I've started by calculating the lift, in vertical autorotation, of the rotor of my gyro for a given angular speed, at first assuming a uniform, mean Cl value across the blade, and then refining that by introducing Cl as a radius-dependent variable. I intend now to add the variable of horizontal speed, and then compute also the drag. In later 'refinements', I plan to add backward disk tilting and flapping... I'm not sure if I'll be capable of that, but I'll try... As a guide, I have the book 'Flugphysik der Tragschrauber', that has a reasonable level. Written for pilots and fans, isn't difficult to understand.

Thanks again for your help...

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#### XXavier

##### Member
Another way of looking at it.

The Predator has a Lycoming IO-320 rated at 160 horsepower and according to the manual burns around .45 gallons per horsepower hour.

The Predator at an all up weight of 1,400 pounds burns around 6 gallons per hour at 45kts and a gallon of gas weighs abound six pounds.

The Predator is burning 36 pounds of fuel per hour divided by .45 and she requires around 80 horsepower to fly at 45kts indicated air speed (52 miles per hour).

Increase the speed to 65kts indicated air speed (75mph) and she burns closer to 7 gallons per hour so she is making around 93 horsepower.

Climb to 7,500 feet density altitude and at 90kts indicated air speed (104 mph) she burns around ten gallons per hour making 133 horsepower based on these simple calculations. I suspect it is a little less because she is using some of the fuel to keep her cool at wide open throttle.

This is around 104kts (120mph) true air speed.

To be sure, you'd need a flowmeter. That instrument would allow you to easily find the airspeed for minimum power required, and once you know that, you know the best range speed too, since (in theory) it's 31% higher...

A nice instrument. A pity that it costs 1400 dollars, if I remember correctly...

#### kolibri282

##### Super Member
Quote: I don't go beyond some simple programs in BASIC /Quote
You could actually do it using a slide rule, Xavier, I am pretty sure the Huey was designed that way. I had compiled a list of reports that might be helpful in the process some time ago, witch you can fined here (post #5):
Rotor Formulae

The key element is calculating the induced inflow for which you can use the incredibly clever diagram 1. in DRA 659. Please tell me if you would like to try it, I would prepare a small example of how to use it next weekend.

Good Luck!

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#### XXavier

##### Member
Quote: I don't go beyond some simple programs in BASIC /Quote
You could actually do it using a slide rule, Xavier, I am pretty sure the Huey was designed that way. I had compiled a list of reports that might be helpful in the process some time ago, witch you can fined here (post #5):
Rotor Formulae

The key element is calculating the induced inflow for which you can use the incredibly clever diagram 1. in DRA 659. Please tell me if you would like to try it, I would prepare a small example of how to use it next weekend.

Good Luck!

Well, I am of the slide rule generation, but shifted to electronic calculators long ago. I use an excellent RPN HP48, discontinued long ago...
Many thanks for the info and the good wishes. I can't, however, to find the 'diagram 1' that you mention...

Concerning the 'induced inflow' (I'm not sure of the meaning of the term, but I believe you mean the net flow striking the rotor blades from below) I calculate it by iteration. I start with a value of 1/3 of the sink speed estimated for autorotation, then compute the lift, compare it with the weight (that should be the same as the lift, since autorotation has a constant descent speed) and then modify the starting value with the ratio weight/computed lift, repeat the calculation with the new value, and so on until I get a good result...

Thanks again...

#### kolibri282

##### Super Member
For a rotor (and the attached helo) to stay aloft the rotor (by Newtons first law) has to accelerate air downward, this is called induced flow.
Diagram (actually fig. 1) is this one

#### XXavier

##### Member
Yes, the 'induced flow' is the 'downwash', and perhaps I'm confused or mistaken, but I term as 'inflow' the air flow striking the (gyro) rotor from below.

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