Head Speed

[scallybert]

What factors affect head speed choice ?

While it's pretty obvious that higher head speed -> greater cyclic & collective effect (similar with tail), what other benefits/disadvantages come with different head speeds ?

[I'd guess that higher headspeeds tend to take more power due to increased drag.]

As an example, the G5 recommends 2200-2300 RPM for beginner/intermediate, 2450 for advanced 3D. I'd been increasing mine (beginner set-up, basic flying) in steps, and inadvertantly ended up around 2550. This was great, and I'm now happy at 2450 (2550 was possibly better - but I was worried about stressing components.)

So, why, as a beginner, might I be better off at 2200-2300 RPM, as the manual suggests ?

My experience suggests that aside from efficiency and component stress, the higher the head speed, the better. (?)

[And that an ESC where you can simply dial-in the desired head speed makes things much easier. But that's a different kettle of fish.]

[Rotorhead]

What factors affect head speed choice ?

While it's pretty obvious that higher head speed -> greater cyclic & collective effect (similar with tail), what other benefits/disadvantages come with different head speeds ?

[I'd guess that higher headspeeds tend to take more power due to increased drag.]

Yes, if you still keep the same cyclic and collective pitch range as before, the drag rises quite quickly, for example if you double the speed you will quadruple the drag force at any particular pitch setting, which quadruples the power requirement for that pitch setting.

As an example, the G5 recommends 2200-2300 RPM for beginner/intermediate, 2450 for advanced 3D. I'd been increasing mine (beginner set-up, basic flying) in steps, and inadvertantly ended up around 2550. This was great, and I'm now happy at 2450 (2550 was possibly better - but I was worried about stressing components.)

Just be careful going over their ratings, 100rpm may not seem like much, and is probably still within the safety margin for the blades, blade grips, blade bolts and spindle, its still going to actually be a fair difference load wise if you where to do the load calculations.

So, why, as a beginner, might I be better off at 2200-2300 RPM, as the manual suggests ?

Its likely they say that because at the lower settings the model will obviously be less twitchy.

My experience suggests that aside from efficiency and component stress, the higher the head speed, the better. (?)

[And that an ESC where you can simply dial-in the desired head speed makes things much easier. But that's a different kettle of fish.]

Better for speed of roll rates etc and climb rates yes, but it will make the model more twitchy when you want to be flying with precision, eg the hover and landing and taking off phases, or anytime you want to slow things down.

[scallybert]

Yes, if you still keep the same cyclic and collective pitch range as before, the drag rises quite quickly, for example if you double the speed you will quadruple the drag force at any particular pitch setting, which quadruples the power requirement for that pitch setting.

Actually, the power would be *8, as power is force*speed or torque*revs; and while force or torque has quadrupled, so has revs or speed.

However, the flaw is assuming the pitch stays the same. To achieve the same lift at higher revs you'll need less pitch.

I suspect we're all out of our depths trying to do this bit theoretically...

Just be careful going over their ratings, 100rpm may not seem like much[...]

You're preaching to the converted. I know the centrifugal forces increase with revs **2, and I didn't intentionally exceed the recommended revs.

Its likely they say that because at the lower settings the model will obviously be less twitchy.

Better for speed of roll rates etc and climb rates yes, but it will make the model more twitchy when you want to be flying with precision, eg the hover and landing and taking off phases, or anytime you want to slow things down.

Hovering, etc is what I'm doing, and I haven't found the G5 twitchy at the higher head speeds (2450 or 2550). If anything, it seems more composed at these speeds.

If the only reason the lower head speeds are suggested is to reduce twitchiness (by making the controls less sensitive), I'd have thought this could have been achieved by twiddling expo/rate/pitch curve/etc.

So far, I can't see why a beginner would want to run at (eg) 2200 rather than 2450.

[brin]

i would say that a beginner is more prone to giving excessive stick commands therefore more pitch at a higher head speed would make it more twitchy.just my thoughts.

[Brian Gifford]

Well, as a beginner, I'm running a comparatively high head speed on my 500, 100% throttle curve on a 13t pinion. It certainly seems less twitchy than lower head speeds to me. Is it possible that there is a gyroscope effect, and the higher the headspeed, the more this effect shows?

Don't know, just thinking aloud.

[brin]

Well, as a beginner, I'm running a comparatively high head speed on my 500, 100% throttle curve on a 13t pinion. It certainly seems less twitchy than lower head speeds to me. Is it possible that there is a gyroscope effect, and the higher the headspeed, the more this effect shows?

Don't know, just thinking aloud.

good point brian i found that with my 450 but i had gone past the t/g and hovering stage by that time.

[Brian Gifford]

good point brian i found that with my 450 but i had gone past the t/g and hovering stage by that time.

Lost the TG, and trying to move past hovering now.

[Rotorhead]

Actually, the power would be *8, as power is force*speed or torque*revs; and while force or torque has quadrupled, so has revs or speed.

I did some similar calculations a while back and a guy called Peter Wales then went on to test the results, and its a lot simpler than that.

If the blade number stays the same, if you double the speed you will quadruple the drag, that is quadruple the force the engine has to produce, which means it needs to produce 4x the power to produce that 4xforce than it did at the original speed.

If you just keep to force its a lot simpler, what ever the drag force is, thats what the engine has to overcome.

Thats based on the drag formula having a V^2 factor in it, if you keep all other things (Drag coefficient, frontal x-sec area, air density) equal in the drag formula then doubling V will quadruple the result.


Like you say the pitch requirement will need to be factored in, in reality frontal x-sec area will decrease because the pitch has decreased or the heli would climb, so in practice it wont be 4x.

Lowering the pitch will also reduce the coefficient of lift and drag (CD and CL)

Or put it another way, if you reduce rotor speed by 50%, then you will only need about a quarter of the power to drive the system even though you need more pitch.

Anyway you're not doubling the speed so its not that big a jump in power requirements.



What Peter Wales wanted to know was how much extra power another blade would require, which is kind of similar to what you want to know.

He wanted to know what the power requirements would be for adding an extra blade so as to have 3, and its a bit counter intuitive, it still goes up even though you have more surface area to produce lift on, at the same headspeed you will generate about 1.4 times the drag since he initially started with 2 blades.

1.5 if you dont factor in the lowered pitch.

Pretty simple, you just find the multiple of the number of blades, 3 being 1.5x2 which gets you close to the answer, just a bit over due to not factoring in the lower pitch, also each time you add a blade the lower the next jump in factor, 1.5 then 1.33, then 1.25 etc

In his test on his electric model on 2 blades he was drawing 20A, and on 3 blades at the same headspeed but lower pitch angle he was drawing 28A

So an estimate of 30 was pretty close.

I suspect we're all out of our depths trying to do this bit theoretically...

Not really, you just need to use the lift and drag formulas to work out the loads, the big problem is figuring out the frontal cross sectional area of the blades at the hover, this is tricky because you have to factor in the downflow speed to calculate the angle of attack, which is going to be less than the angle of pitch because of the resultant vector of the airflow.

Just need a wind tunnel (sure I've got one of them somewhere arround here) or some source of data on how fast the flow is through the disk.

If the only reason the lower head speeds are suggested is to reduce twitchiness (by making the controls less sensitive), I'd have thought this could have been achieved by twiddling expo/rate/pitch curve/etc.

Yes.

[scallybert]

If the blade number stays the same, if you double the speed you will quadruple the drag, that is quadruple the force the engine has to produce, which means it needs to produce 4x the power to produce that 4xforce than it did at the original speed.

I didn't disagree about 4* the torque.

However, you've also doubled the revs.

4* the torque and 2* the revs gives 8* the power.

[Rotorhead]

Ah right I see what you're saying.

Agreed.

[Gareth-71]

Hey, just my two cents

Basic aerodynamics dicates that drag is the square of speed, so the effect of higher speed is squared. However there is also mechanical efficiency, which is a small factor as everything runs on bearings, so I would discount it in the calculations. Also, the efficiency of the motor is a sizable factor, but BDC motors tend to be more effiecient at higher RPM (within reason).
So the overriding factor is aerodynamic drag on the blades, which is simply the sqaure (not quadruple which implies multiply by 4, unless the speed is 2)

That is why speed records on bikes and cars is so hard, just calculate the extra power you would need to increase top speed from 200MPH to 220MPH.

i.e. You want to increase the speed by 10%, which means that you will need to increase the power by 100% !!!, i.e. you have to double the power at that speed to get a 10% speed increase.


HOWEVER------

Head Speed, or head RPM is directly proportional to blade tip speed, so the same is true for head RPM, HOWEVER, things get a lot more complex because....The motor power output for a given head speed is X to, say, maintain a hover, if you increased the speed, you would equally decrease the pitch to achieve the same hover, i.e. you would decrease aerodynamic drag when you increase head speed to maintain the same load.

Drag is therefore not a constant it depends on pitch.

So the loss in efficiency is the difference between the two, which is very, very difficult to calculate (because it is not a constant you would need to calculate the difference in drag between the two).
Far easier to hover a pack at low head speed and time it, and then the same at high head speed and time it.


It will probably work out quadruple! :-)

Gareth

[Rotorhead]

The speed is a square, thats why I said if you double the speed you quadruple the drag, work it out.

If you say start with a speed of 2m/s and everything else in the drag formula is 1 (for ease of the example) then the drag force will be 4 newtons, if you double that speed to 4m/s then the drag force will be 16 newtons.

That is four times (quadruple) the original drag force, and that works for other speeds.

What ever the drag force is at speed x, if nothing else in the equasion changes, if you double that speed then the new force is 4x the old one.

Head Speed, or head RPM is directly proportional to blade tip speed, so the same is true for head RPM, HOWEVER, things get a lot more complex because....The motor power output for a given head speed is X to, say, maintain a hover, if you increased the speed, you would equally decrease the pitch to achieve the same hover, i.e. you would decrease aerodynamic drag when you increase head speed to maintain the same load.

Drag is therefore not a constant it depends on pitch.

I believe I covered all that.....

Please read more carefully.

[Gareth-71]

Hi!

Sorry to be a stickler, but quadruple is not square.

2 sqrd = 4
3 sqrd = 9
4 sqrd = 16 (quadruple)
5 sqrd = 25
....
2 quadrupled = 8
3 quadrupled = 12
4 quadrupled = 16 (same as squared)
5 quadrupled = 20
....

[Rotorhead]

I was on about the answer and I was very clear about that when I said it.

I was specific about it being the case of doubling your speed.

If your first answer is 2 and the second answer is 8, then even though its a square rule that made the difference the answer has gotten four times bigger.

It really does ammount to the same thing in formulas like the lift and drag formulas and it demonstates the point that the formula is dominated by a square law.

[Gareth-71]

Eh?

Dude, I am now totally confused by what you are saying...

Anyway, I think everyone has got the point. If you want to argue the finer points of mathematics, great, but do it via PM.