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Old 08-11-2010, 04:31 PM   #1 (permalink)
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Lightbulb FYI: the relationship between headspeed and flight time

Some of you might know that I experimented a bit lately with different pinions and headspeeds in governor mode on the Furion, and at each different flight I measured the headspeed and the capacity that the charger needed to charge the packs completely again. With that, I calculated the possible flight time for a 80% discharge of the packs. The used setup was a stretched V-Barred Furion 450 with a Scorpion 2221-1630kV, 13 and 15t pinion, Jazz ESC, two 3S/1600mAh/20C packs in series and 350 mm blades (bolt hole to blade tip).

All these data gave me some insight in the relationship between headspeed and flight times for this particular heli. But even if the conclusions are only correct for this heli, there are possibilities to extend this to other 450s and even bigger helis later on, with enough extra measurements of course. Might be an idea for an extra tool in HeliPort, BTW.

Anyway, I found it interesting to do (yes, I'm such a nerd...), and I thought I would share this, hoping it might be of some use anyhow. Here’s a screenshot of the spreadsheet that I made with the data:




The bottom part of the graph is interesting, but of no real practical use on 450s needing relatively high headspeeds: the flight times increase dramatically with very low headspeeds. There seems to be a linear relationship for the rest of the headspeeds. Much depends on the flying style, but I tried to keep it as consistent as possible troughout the different flights. There are enough samples to show clearly that the relation is consistent enough to make some conclusions.

A smaller (13t) or larger (15t) pinion, when used in such a way that they generated the same headspeeds (different motorspeeds), didn’t seem to change much in the total efficiency. You can see the red (15t) and blue (13t) curve nicely overlapping each other. The Jazz is known to be very good in handling extremely low throttle percentages without heating up because of extra dissipation for example as some other ESCs do, so no extra loss of energy.

This also teached me that (still only for this heli), every 100 RPM less means about 35-40 seconds of extra flighttime at least ! And with a bit of math, it was possible to make the equation: flight time in seconds = -0.375 X RPM +1616. Meaning that if you know the RPM that you like , and set up the heli for that value, you also know the flighttime at once, without any testflights.

Well, this might not be of much use for now, although I suspect that the inclination of the curve might be rather similar for other 450 sized helis. I might check this with the Beam later on. But if one would add the weight of the heli, blade size and battery capacity to the equation, one would have a way of predicting the approximate flighttimes for almost any choosen setup, at least for 2 bladed electric helis. That would mean retrieving a lot more data first, of course. It might be done already, I even didn’t check on that yet.

The most obvious conclusion is that higher headspeeds really eat a whole lot of battery capacity, so fly at the lowest RPMs that you really want or need for your setup and flying style, all the rest seems wasted energy, battery usage and practice time. But the fun factor is also important in this hobby of course. Just my 2 eurocents for now…
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Old 08-11-2010, 05:52 PM   #2 (permalink)
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Awesome work there Raf. It is obvious you have put an awful lot of effort into that. It will be great to see how this develops as you factor in weight and other variables. I look forward to further reports.

Cheers

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Old 08-11-2010, 06:55 PM   #3 (permalink)
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Good job, Raf! Looks like around 3100 rpm is the head speed to shoot for. According to what I see in you graph, very nice by the way, anything above that consumes more energy that it's worth.
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Old 08-11-2010, 09:37 PM   #4 (permalink)
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The power goes up with the cube of RPM. Extra head speed is a very expensive component of the power budget. The heli will fly longer in forward flight than at hover (out of ground effect) because the forward velocity adds lift that lets it maintain lift with a lower headspeed.
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Old 08-12-2010, 05:12 AM   #5 (permalink)
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Indeed. And one more consideration is how much of this energy profit (because of the extra lift in forward flight) would be lost by the fact that one needs to apply extra collective to keep it in the air at the same altitude while the nose is pointing slightly down, with a smaller horizontally projected area of the blade disc. Anyway, would be nice to check the difference in power consumption between a hover and FF, and that’s easy to do in fact.

Another thing: adding the battery capacity to the equation is very simple, if you would neglect the difference in weight of the packs in first instance. Simply multiply by the used capacity, and divide by the reference capacity, being 1600 mAh in this case. But you also need to take the weight in consideration, and that would involve another series of tests and measuring discharged capacity using the same packs, and adding for example 50 grams more at each flight, from 50 up to 500 grams or so. That would allow to extend the equation for this size of heli with different weight and different packs. I’m thinking that there is a way to make a universal formulae for most 450 sized helis using the same blade sizes. As well as for other sizes of helis. But combining all these to one general formulae that could be used for all sizes could be very hard, too many variables are changing at that moment. Well, we’ll see where we get, don’t expect anything soon though, lol.

Edit: a bunch of other stuff would need to be considered also: like flybarred or not, torque tube or not, etc...
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Old 08-12-2010, 09:30 AM   #6 (permalink)
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I gave this one some further thoughts, and at first sight it looks like flight time can be calculated theoretically with an equation of this form:

Flight time = a function of (headspeed; total weight; battery capacity; blade size and profile; efficiency of motor and gears [inrunner/outrunner/efficiency rate/…]; tail drive system [TT/belt]; head system [flybar/flybarless]; other heat and friction losses; aerodynamic drag; power consumed by the electronics; flying style)

To make it possible to do some real math, we need to simplify things, without making things very inaccurate (hopefully):

Let’s say we aim for one formulae for each type of heli, this would make the variable blade size unnecessary, and drag and blade profile won’t make that much difference, so leave that one out also for now.
We first need to take a reference for flying style, like simply hovering, and subtract some flight time depending on how hard the pilot flew, but that can easily be measured in real life.

By working from real life measurements, we could also leave out the heat and friction losses, and assume them equal for every 450 for example, hence cancelling each other in the math. Just like the power needed for the electronics, the flight times are too short and the currents too low to play an important role.

Most other unknown variables can be found when doing enough test measurements with different setups, where only one variable at a time has been changed, and compare those between them. The differences in efficiency of the motor and drive system would be the most difficult matter.
Make a formulae out of this, use it in a software algorithm, and you’re done. Should take only a few years at most, lol.
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Old 08-12-2010, 11:00 AM   #7 (permalink)
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So if the math is correct a "reference standard" is born,
for example :

If someone has a belt cp and he can't make the calulated flighttime within a margin there is something wrong like the belt to tight or bad batts or wrong rpm's.
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Old 08-12-2010, 04:00 PM   #8 (permalink)
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Quote:
Originally Posted by RocketSled View Post
The power goes up with the cube of RPM.
No it doesn't, RS. It's rather lineal, as Raf's experiment shows. I'm no expert on fluid dynamics, but in the normal (non-turbulent) range it seems quite lineal, probably due to the fact that the torque load requirement on the blades (resulting from the blade's pitch) decreases as the head speed goes up. So I think that the power relation to RPM is actually square (with constant pitch), but the pitch decreases in a linear manner so the net power effect is lineal increment of the power consumption (and decrease of the flight time).
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Old 08-14-2010, 01:32 PM   #9 (permalink)
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Well, first I had to go get a coffee to sip while I pondered this.

As usual Raf, outstanding work

I always look forward to your work to keep my old brain fully engaged
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Old 08-14-2010, 05:54 PM   #10 (permalink)
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Quote:
Originally Posted by jperkosk View Post
No it doesn't, RS. It's rather lineal
Yeah, no, that's not how I understand it. The power required to turn a propeller is governed by the equation:
  • shaft power = k * RPM^3 * diameter^4 * pitch

Here's why power increases with the cube of RPM:

Power = Torque * RPM / 5252.

Every place I use "=" substitute the word "proportional to". So...

p = t * RPM

The drag of a wing increases with the square of velocity. For a rotor blade, the velocity is proportional to the RPM. Since velocity is proportional to RPM, we can "loosely" (meaning, ignoring the other terms and constants) substitute RPM for velocity, and the blade's drag becomes proportional to the square of its RPM instead of velocity. The equation becomes:

d = RPM^2

The power system must generate enough torque at the main shaft to balance the rotor's drag in order to keep the rotor at a constant speed. So shaft torque is directly proportional to drag. The equation is:

t = d

Since drag is proportional to the square of RPM, that means torque is, too. Substituting for "d", the equation becomes:

t = RPM^2

Power is proportional to torque times RPM, from above...

p= t * RPM

Substituting for "t", the equation becomes:

p = RPM^2 * RPM

So Power must be proportional to the cube of RPM.

p = RPM^3
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Old 08-15-2010, 03:33 AM   #11 (permalink)
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RocketSled, you made me strain my brain and look up fluid dynamics on Wikipedia ...

OK, looks like your equations are all correct for very high speeds (where flow is turbulent), which is referred to as high Reynolds number type of the flow.

http://en.wikipedia.org/wiki/Drag_%28physics%29

Quote:
Originally Posted by Wikipedia
Reynolds numbers frequently arise when performing dimensional analysis of fluid dynamics problems, and as such can be used to determine dynamic similitude between different experimental cases. They are also used to characterize different flow regimes, such as laminar or turbulent flow: laminar flow occurs at low Reynolds numbers, where viscous forces are dominant, and is characterized by smooth, constant fluid motion, while turbulent flow occurs at high Reynolds numbers and is dominated by inertial forces, which tend to produce random eddies, vortices and other flow instabilities.
I think an airplane propeller will work at this range at 15,000 - 20,000 RPM while the helicopter blades operate closer to 2,000-2,500 RPM which I think puts them in laminar flow range (no turbulence).

Now, as you've said:

Quote:
Originally Posted by RocketSled View Post
Every place I use "=" substitute the word "proportional to". So...

p = t * RPM

The drag of a wing increases with the square of velocity. For a rotor blade, the velocity is proportional to the RPM. Since velocity is proportional to RPM, we can "loosely" (meaning, ignoring the other terms and constants) substitute RPM for velocity, and the blade's drag becomes proportional to the square of its RPM instead of velocity. The equation becomes:

d = RPM^2

The power system must generate enough torque at the main shaft to balance the rotor's drag in order to keep the rotor at a constant speed. So shaft torque is directly proportional to drag. The equation is:

t = d

Since drag is proportional to the square of RPM, that means torque is, too. Substituting for "d", the equation becomes:

t = RPM^2

Power is proportional to torque times RPM, from above...

p= t * RPM

Substituting for "t", the equation becomes:

p = RPM^2 * RPM

So Power must be proportional to the cube of RPM.

p = RPM^3

In laminar flow range (where I think helicopter blades are operating) drag would be represented by the linear dependency:

d = RPM

This, however, holds true only for the heli bolted down (that's why we see very quick battery drain on the lazy susan, much higher than while flying). In the hovering scenario as your RPM increases you need to reduce the blade pitch (and blades' frontal area A with it), otherwise your bird will shoot up. As the lineal drag equals frontal area multiplied by speed, I hold that in hovering heli the following is true:

d = A * RPM = constant

Therefore, what follows is:

d = constant
t = d
p = t * RPM
p = RPM

Which kind of explains why Raf's experiment has shown the battery flight time to be linearly dependent (inversely proportional) on the head's RPM.
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Old 08-15-2010, 03:37 AM   #12 (permalink)
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Guys, you are both great ! I love these kinds of discussions, lol.
Remember, it's not really about who is right or wrong, it's all about the learning proces.
Keep them coming !
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Old 08-15-2010, 03:53 AM   #13 (permalink)
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Quote:
Originally Posted by redbird300 View Post
Guys, you are both great ! I love these kinds of discussions, lol.
Remember, it's not really about who is right or wrong, it's all about the learning proces.
Keep them coming !
It keeps the brain exercised, so maybe we can delay the onset of Alzheimer's, too!
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Old 08-15-2010, 11:35 AM   #14 (permalink)
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Alas, my math skills are not up to the task at hand. We've progressed beyond the point where my faculties are equipped for further debate.

I can tell you however that in my past employment life I made Enterprise-class HDDs (10K/15K RPM, SCSI/SAS/FC, if you know what these things are). The power-goes-up-with-the-cube-of-RPM applies to the flat disk inside an HDD, which has the lowest possible cross-sectional drag profile (in fact, it's zero since there is no "leading edge" or angle of attack) and which develops an attached flow at the surface of the disk which is quite laminar (otherwise the recording heads wouldn't have an "air bearing" to fly on).

I agree that as RPM increases, lift increases as well, and this allows you to reduce pitch which reduces drag. I don't know how airfoil drag relates to angle of attack, but I suppose if there's an exponential relationship there the reduction in pitch will at least partially offset the cubed RPM term in the overall power equation... Perhaps I'll do some additional study and find out.

But that all being said, at a 0ş AoA, a fully symmetric heli blade is generating no lift, but it still has aerodynamic drag. That drag still increases with the square of RPM (=velocity), even though the AoA doesn't change as RPM increases, and that means power still has to increase with the cube of RPM per my overly-simplistic analysis from a few posts ago.

Clearly, this is a much more complicated topic than we'd like it to be!
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Old 08-15-2010, 12:32 PM   #15 (permalink)
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Great discussion guys!
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Old 08-15-2010, 03:29 PM   #16 (permalink)
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RocketSled, you are absolutely correct. I did some sample calculations based on a couple of formulas I Googled and they disproved my idea that drag increases linearly with the speed in the helicopter RPM range, to the contrary, we're smack in the middle of the standard airflow range.

So we're back to to aerodynamics 101, and let's start here:

http://wright.nasa.gov/airplane/drageq.html

D = 0.5 * Cd * r * V^2 * Ar
Where
(D) drag
(Cd)
drag coefficient
(r)
density of the air
(V) velocity
(Ar)
reference area (frontal area)

You are correct in that all things being constant d = RMP^2 but we know that A is being reduced with the pitch, the question remains how much. To answer that we need to find out what happens to lift when we increase the RPM, because we need to maintain constant lift to keep the heli hovering, so back to our friends at NASA:

http://wright.nasa.gov/airplane/lifteq.html

L = .5 * Cl * r * V^2 * A
Where
(L) lift
(Cl)
lift coefficient
(r)
density of the air
(V) velocity
(Aw)
wing area

Again, as per NASA, for thin airfoils, at small angles of attack, the lift coefficient is approximately two times pi (3.14159) times the angle of attack expressed in radians.
Cl = 2 * pi * angle (in radians)

After substituting Cl we get:
L = pi * angle (in radians) * r * V^2 * Aw

To maintain lift constant at increasing head speed the angle of attack has to decrease in the square proportion to the speed, meaning (I'll skip all the constants from now on):
angle = 1 / (V^2)

This reduces the frontal area Ar in the same proportion, so without the constants:
Ar = 1 / (V^2)

Substituting Ar in the drag equation (again, no constants):
D = Cd * r * V^2 * Ar = Cd * r * V^2 * (1 / (V^2) = Cd * r = constant

Conclusion: drag increases with the square of the speed for a given pitch of the blades, but so does the lift (kind of logical if we think about it, it's the same force of the airflow over the wing, one is just being at 90° to another). In fact, the drag and lift equations are exactly the same! In other words the pilot of the hovering heli when the head RPM changes manipulates the blade pitch to maintain lift (and drag) constant.

Which of course leads to P = RPM which Raf has proven at the beginning of this thread beyond any doubt , we're just looking for mathematical explanation.

One last question: Raf, you didn't happen to have a data logger on your ESC during the experiment, did you? As the drag at hovering is constant regardless of the head RPM, and in permanent magnet motors torque is th
e same as current, it would show that hovering Amps were always the same regardless of the head speed.

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Old 08-15-2010, 04:12 PM   #17 (permalink)
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Quote:
Originally Posted by RocketSled View Post
I can tell you however that in my past employment life I made Enterprise-class HDDs (10K/15K RPM, SCSI/SAS/FC, if you know what these things are). The power-goes-up-with-the-cube-of-RPM applies to the flat disk inside an HDD, which has the lowest possible cross-sectional drag profile (in fact, it's zero since there is no "leading edge" or angle of attack) and which develops an attached flow at the surface of the disk which is quite laminar (otherwise the recording heads wouldn't have an "air bearing" to fly on).
Yep, it makes sense. In this scenario Ar and Aw never change, so the drag and lift go up with the square of the RPM, and the power with the cube of RPM.

Quote:
Originally Posted by RocketSled View Post
Clearly, this is a much more complicated topic than we'd like it to be!
It's for the scientists to complicate the heck out of this world and leave it up to us engineers to bring it down to earth and make sense out of it.
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Old 08-15-2010, 04:17 PM   #18 (permalink)
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Well, I’m usually not easily impressed, but the fine structure, clarity, well-documented and logical posts that you both made here makes me declare publicly that I’m very impressed, gentlemen. This is rarely seen on a r/c heli forum, and feels more like being worthy to be published on a scientific forum.

It all made good sense when simply reading it, but I know from experience that it takes a lot of persistency and brain-boiling tinkering before you get to such conclusions if starting from a situation where you don’t have all the necessary knowledge right at your disposal. I recently lived the same situation when learning more about Fast Fourier Transformations and sound waves physics while I was writing the audio tachometer in HeliPort, which was rather new for me, and I'm even not a professional programmer. Intelligence alone doesn’t cut it at that moment, hard work and logical thinking combined with the right mindset is far more important. So, congratulations. Something tells me that we’re only at the beginning of this story, there are some unanswered questions when reading between the lines, but nevertheless, EXCELLENT JOB !


Now Jerry, unluckily I did NOT have a data logger when doing the experiment, but if I would find the time, I will repeat all the tests and more with the Beam, which has a CC ICE ESC with logging features.

BTW, just a small detail: while the whole of your post is very clear, I’m still wondering about your very last line, maybe I didn’t understand what you really mean, I’ll give it some thoughts. The experiment showed that average hovering amp draw is higher at higher RPMs, and I can’t seem to combine this with that last line ? Forgive me for asking, it does not take anything away from your achievements.
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Old 08-15-2010, 04:57 PM   #19 (permalink)
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Quote:
Originally Posted by redbird300 View Post
BTW, just a small detail: while the whole of your post is very clear, I’m still wondering about your very last line, maybe I didn’t understand what you really mean, I’ll give it some thoughts. The experiment showed that average hovering amp draw is higher at higher RPMs, and I can’t seem to combine this with that last line ? Forgive me for asking, it does not take anything away from your achievements.
Hi Raf, thanks for the good word, just trying to delay this Alzheimer's onset thing here

This one is actually easy as I'm an elecrtical / automation engineer and deal with motors, torques and currents daily:

On the permanent magnet motor side the current means torque and the voltage means speed, period. Drag is constant so the current should remain constant as well. I was referring to motor current. The increased power consumption (P = RPM in our case) will be reflected in the voltage, as electrically P = V * I

On the battery side the voltage supply is constant (ignoring the discharge curve), so what ESC is doing is drawing more current when it needs to produce higher voltage to the motor. All brushless ESCs are of the switching type (I think), so they regulating motor voltage by connecting the battery to the motor leads (at high frequency, few kHz) only a percentage of the time.

Example, let's say our battery is @ 10V under the load. ESC switching it on 40% of the time will produce 4V output and 2500k motor will run & 10,000 RPM. The same motor with ESC @ 80% will see 8V applied to it, will run at 20,000 RPM, but the motor current remains the same as you're still hovering the same heli, so the drag is the same, right?
The battery, however, will see the motor leads connected to it 80% vs. 40% of the time, so the current drain from the battery will be double.
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Old 08-15-2010, 05:06 PM   #20 (permalink)
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OK now, you referred to the current between the ESC and the motor, not between the pack and the ESC. I'm an electronics guy myself, the way a switching speed controller works is familiar to me, all clear now. I already thought that this one was far too obvious, but I couldn't make it up in that last line, sorry for the misunderstanding.
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