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Old 10-02-2011, 09:37 AM   #21
airjawed
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All excellent points here from everyone. I don't think anyone is disputing all the effects and variables at play on a heli. In fact, this is definitely a big part of what makes this hobby so much fun for me.

Anyone have a data logger for power consumption and altitude meter on the heli? You could get some info on overall power used vs power done. Compare upright vs inverted climbouts and that should be a very good comparison of aerodynamic efficiency.
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Old 10-02-2011, 06:56 PM   #22
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Originally Posted by wlfk View Post
If you look at the tail rotor, the majority of them push the air away from the tailfin rather than towards it, as it's more efficient this way round - air sucked into the tail rotor comes from a larger area than the jet of air blown away from it.

In the same way, air blowing away from the main rotor has a higher velocity than air sucked towards it, so although airflow over the shape of the canopy may be more efficient in one direction rather than another, we would also expect the helicopter to fly more efficiently if the airflow from the rotor does not have to pass over the canopy.

Wlfk

There are 10s of variables involved in the question, most of which interact. But I picked the canopy question because it is easier for most people to understand, and they seem to focus on it as a cause.

The data I presented took some effort to calculate. If you have evidence to suggest the finding is incorrect, Id like to see it.

The reality is, air flowing over a stock canopy from the bottom (i.e. in an inverted hover) creates more drag than it does when it flows over the top of the canopy (upright hover). In a hover, the airspeed over the canopy in upright hover is approx 2x higher than inverted, and drag is a function of airspeed squared. So, if you do the math, inverted hover for a stock canopy is more efficient; it has a higher drag coefficient, but is seeing slower air. But as the machine accelerates inverted, the drag advantage drops rapidly, because the difference is airspeed is no longer 2/1. By the time you reach 8.1mph, the airspeed ratio is 1.45/1, which is where the drag becomes a break-even. Above that speed, the drag becomes a net negative inverted.

Just an sort of an interesting visual, have a look at this vid:


Aside from it being a cool video with a great soundtrack and a very smooth pilot, there are sections where the canopy has collected rain. You will see in those sections the surface flow pattern over the canopy. Look at 3:30 for example, and on departure, you will see how dramatically the water droplets change from a vertical path to horizontal as he accelerates. That shows you the velocity components in play- where even at a forward airspeed of 10-15 mph, the droplets are nearly horizontal. There is just not that much wash on the canopy, and it is far less for our models, which operate MUCH lower disc loading.
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Old 10-02-2011, 07:40 PM   #23
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Originally Posted by airjawed View Post
All excellent points here from everyone. I don't think anyone is disputing all the effects and variables at play on a heli. In fact, this is definitely a big part of what makes this hobby so much fun for me.

Anyone have a data logger for power consumption and altitude meter on the heli? You could get some info on overall power used vs power done. Compare upright vs inverted climbouts and that should be a very good comparison of aerodynamic efficiency.

Well, the original question is 'which is faster', not which is more efficient. Not necessarily the same I do have the hardware for testing all the rates/power levels, but without a tunnel, it would not be possible to eliminate stuff like head asymmetry.

This really needs a wind tunnel with proper high speed cams and all that...
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Old 10-02-2011, 09:37 PM   #24
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True the original was 'which is faster' but that was under the presumption that the head was setup symmetrically and slop was not a factor. I'm 100% sure that there are all the other variables at play such as described by yourself and others.

However, when it comes to 250 size helis I feel it is a slop/asymmetrical pitch issue. Then, when it comes to pros it is the effect of ground push-off and the good old 'cool factor'... IMHO

On a small heli when pitch is setup using a gauge the main shaft/head is pushed downwards which biases the slop to falsely show additional positive pitch. Additionally, the pitch gauge is typically not balanced on small blades well. This buts weight typically on the trailing edge of the blades.... biasing the pitch additionally to the positive side. End result is the in reality if for example you think you have a -10,0,+10 setup you in actuality have -11,-1,+9..... this makes for a faster inverted climbout.
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Old 10-03-2011, 12:04 AM   #25
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Ground effect exists practically only for a rotor diameter’s distance from the ground, and even then, only adds about 15% efficiency. The rotor diameter/ground effect rule is independent of upright/inverted; it is simply disc to ground.
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Old 10-03-2011, 09:03 AM   #26
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Sutty-

Of the potential subsystems, look at the one (canopy drag) people intuitively seem to hold as primarily responsible for the reported inverted advantage-

If you do the math, and assuming mine is correct, you will find that the flow velocity in question is low; average flow velocity change through a stock 450 rotor system at max power is approx 17 fps, or about 11.5mph. Half the acceleration happens on the inflow side, half on the outflow side. So, inverted, same scenario, the canopy sees 8.5fps.
On what basis do you claim that half the acceleration happens on the inflow side and half on the outflow side?

If you imagine a ducted fan, you'll find that the airflow in front of the fan and the airflow behind it aren't terribly different. It doesn't make sense to talk about half the acceleration happening in front of the fan, and half behind. There'll be a small difference due to compressibility, but it won't be anything like 1/2 ratio in air velocities.

The reason a rotor has a higher velocity beneath the rotorblades is that, if you imagine it as a 'virtual' ducted fan, the shape of the duct would be more 'conical' with the air outlet significantly smaller than the air inlet. In other words, a larger volume of air is being accelerated to a lower speed above the rotor, and a lower volume of air is being accelerated to a higher speed below the rotor.

There's a figure illustrating this for tail rotors on Leishman, pg 318:
http://books.google.com/books?id=nMV...pusher&f=false

The difference in efficiency for a pusher or tractor tail rotor seems to be about 6%, but this depends partly on the area of the tailfin, so obviously won't translate directly to the performance of a main rotor.

And another figure showing the vena contracta here, more clearly:

http://www.aerospaceweb.org/question/aerodynamics/rotor/rotor.jpg


Quote:
Lookup the coefficient of drag for a 450 canopy in the Z axis, or measure it yourself. It produces about 2x the drag for a given air velocity coming at it from its base vs from its top (Cd is approx 2x higher for negative Z flow).
Where do you look up the vertical coefficient of drag for a 450 canopy?

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The point of this is- the magnitude of form drag delta in question- including the boom, the gear, tail fin, etc, is generally quite small (< 5%) at the speeds we are talking about (Z-axis climbouts). And, in some cases, advantage becomes disadvantage as the system state changes. If the margin is substantially higher than 5%, probably makes sense to look at other subystems (tail rotor seems high on that list)
Don't disagree with looking at other factors such as the tail rotor, but I'm uncertain about some of your working.
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Old 10-03-2011, 04:56 PM   #27
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You can get the rotor closer to the ground when inverted... no heli frame getting in the way.

If anyone wants to do a simple experiment to see how much slop there is:

Put a pitch gauge on a 250 or other small heli. See what pitch you have upright... then simple hold heli upside down and see how off the gauge is.... it'll easily be more than 1 degree on a small heli.
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Old 10-03-2011, 07:18 PM   #28
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Originally Posted by airjawed View Post
Put a pitch gauge on a 250 or other small heli. See what pitch you have upright... then simple hold heli upside down and see how off the gauge is.... it'll easily be more than 1 degree on a small heli.
This would only work if you had the blades spinning at flight RPM, as you need the load to see where everything will sit. And I dont know about your 250 but if my 200 had that much slop in the links they would be needing replacement. Working the blade grips by hand i can barely move them at all, so not much slop there.
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Old 10-03-2011, 07:51 PM   #29
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True you need to be in flight to truly see what is up. I don't have a 250... but on my 450 I can get maybe 0.5 degrees of movement by hand. 250's I've seen generally are sloppy ... you see it in the tail as well.
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Old 10-04-2011, 03:10 AM   #30
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On what basis do you claim that half the acceleration happens on the inflow side and half on the outflow side?
Rotor physics is the basis. Here is a starting point: http://web.mit.edu/16.unified/www/SP...es/node86.html


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Originally Posted by wlfk View Post
If you imagine a ducted fan, you'll find that the airflow in front of the fan and the airflow behind it aren't terribly different. It doesn't make sense to talk about half the acceleration happening in front of the fan, and half behind. There'll be a small difference due to compressibility, but it won't be anything like 1/2 ratio in air velocities.
Fortunately, physics doesn’t depend on what makes sense to you-


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Originally Posted by wlfk View Post
The reason a rotor has a higher velocity beneath the rotorblades is that, if you imagine it as a 'virtual' ducted fan, the shape of the duct would be more 'conical' with the air outlet significantly smaller than the air inlet. In other words, a larger volume of air is being accelerated to a lower speed above the rotor, and a lower volume of air is being accelerated to a higher speed below the rotor.
You are confusing cause and effect. The rotor imparts energy to the airstream; total pressure is increased, where dynamic pressure goes up (velocity) and static pressure changes very little. As you will learn from the link above, half the acceleration happens prior to the disc, half after. The so-called vena contracta (wake contraction) is the result of conservation of mass; it is Bernoulli at work. At any cross section perpendicular to the flow stream, the mass flow must remain constant. Because the velocity in the far wake is 2x the inflow, its volume will be about half that of the rotor. That would be 1/sqrt(2), or about .707 the diameter of the rotor. It is simple physics mate.


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Where do you look up the vertical coefficient of drag for a 450 canopy?
For Cd, you will need to find a reference. Mine is simulated in CFD and confirmed in a tunnel. But that is just for my canopy- there could be massive deltas depending on the design, if there are holes under the canopy, etc.
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Old 10-04-2011, 05:45 PM   #31
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Rotor physics is the basis. Here is a starting point: http://web.mit.edu/16.unified/www/SP...es/node86.html

Fortunately, physics doesn’t depend on what makes sense to you-

You are confusing cause and effect. The rotor imparts energy to the airstream; total pressure is increased, where dynamic pressure goes up (velocity) and static pressure changes very little. As you will learn from the link above, half the acceleration happens prior to the disc, half after. The so-called vena contracta (wake contraction) is the result of conservation of mass; it is Bernoulli at work.
OK, I think I see where you're coming from. And it was my bad to question your statement on acceleration. However, the drag of the canopy depends on the velocity of the air as it passes it. If you look at a single air molecule as it's sucked through the rotor, it isn't immediately accelerated from v to 2v as it passes through the rotor. The process is gradual as the sigmoid curve in figure 11.26 (top) of your link shows.

Let's say you have two otherwise identical helicopters, one with a rotor mast of 10cm and one with a rotor mast of 20cm... Hover them first of all upright, then inverted, and measure the velocity of the air passing the canopy in each case... The ratio will be higher for the 20cm mast than for the 10cm mast, therefore it's not a constant. To take the argument further, imagine measuring the velocity immediately above and below the rotor... It will be identical (because acceleration, by definition, is not instantaneous in either time or space).

This is why I'm arguing that you can't be certain that the ratio will be 1:2 without doing a lot more work or simulation, than simply arguing that half the acceleration happens above the disc, and half below.

Quote:
At any cross section perpendicular to the flow stream, the mass flow must remain constant. Because the velocity in the far wake is 2x the inflow, its volume will be about half that of the rotor. That would be 1/sqrt(2), or about .707 the diameter of the rotor. It is simple physics mate.
I think you mean area?

I'm fairly confident in arguing that the ratio of the airspeed passing the canopy when inverted:upright is not fixed. However, your argument about CdA would still make sense if the velocity ratio can't exceed a particular limit of 1:2.

I can see that the ratio of the vena contracta to the area of the rotor disc may be fixed and relatively calculable. However, I don't immediately see why there should be a similar lower limit on the velocity of the air above the rotor disc? At some distance, it will sooner or later approach zero (this isn't apparent on your link, presumably because it's talking about a propeller that will be moving through still air).

In other words, I don't immediately see that the ratio can't be larger than 1:2, though I'm open to education and I'll admit I would have probably guesstimated the ratio to be less.

Quote:
For Cd, you will need to find a reference. Mine is simulated in CFD and confirmed in a tunnel. But that is just for my canopy- there could be massive deltas depending on the design, if there are holes under the canopy, etc.
I would be very surprised if there are references where you can just look this up. Working it out homebrew, as you have, seems to me to be the only way (unless some kind hearted soul has posted their measurements, or unless perhaps you work for Align!).
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Old 10-04-2011, 08:02 PM   #32
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OK, I think I see where you're coming from. And it was my bad to question your statement on acceleration. However, the drag of the canopy depends on the velocity of the air as it passes it. If you look at a single air molecule as it's sucked through the rotor, it isn't immediately accelerated from v to 2v as it passes through the rotor. The process is gradual as the sigmoid curve in figure 11.26 (top) of your link shows.

Let's say you have two otherwise identical helicopters, one with a rotor mast of 10cm and one with a rotor mast of 20cm... Hover them first of all upright, then inverted, and measure the velocity of the air passing the canopy in each case... The ratio will be higher for the 20cm mast than for the 10cm mast, therefore it's not a constant. To take the argument further, imagine measuring the velocity immediately above and below the rotor... It will be identical (because acceleration, by definition, is not instantaneous in either time or space).

This is why I'm arguing that you can't be certain that the ratio will be 1:2 without doing a lot more work or simulation, than simply arguing that half the acceleration happens above the disc, and half below.

The wake contraction happens some distance from the rotor plane because air is still accelerating. So, while for a 450 rotor (call it 28” diameter) the inflow velocity 10cm above the disc on the inflow side is not quite X, the outflow velocity 10cm below the rotor is not quite 2X.

One can postulate all kinds of arguments about these flows. For example, inflow is non-directional, and outflow is. Conversely, the upright variation has much of the canopy within the grip area, where very little acceleration happens.

The ratio will likely not be 2:1- it is very dependent on the proximity of the canopy to the rotor, and the rotor diameter, etc. None of this is particularly specific- it can't be, since there are huge variations in machine design, head speeds, etc.

Even if it is 3:1, the breakeven on drag increases to just 15mph. Above that speed, the inverted canopy is more draggy. By the time the machine reaches 27mph, the inverted canopy is generating 30% more drag than it would upright in the wash.

Keep in mind, for benefit of the doubt my calcs are assuming the canopy is seeing the entire average velocity (2X). It isn’t- it is seeing less than half that due to its position in the slower velocity flow core. In that scenario, the breakeven (even assuming a 3:1 flow delta) is at 6.8mph…



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Originally Posted by wlfk View Post
I think you mean area?
Yes- I meant area, was typing too fast. Thanks for the correction


Quote:
Originally Posted by wlfk View Post
I'm fairly confident in arguing that the ratio of the airspeed passing the canopy when inverted:upright is not fixed. However, your argument about CdA would still make sense if the velocity ratio can't exceed a particular limit of 1:2.

I can see that the ratio of the vena contracta to the area of the rotor disc may be fixed and relatively calculable. However, I don't immediately see why there should be a similar lower limit on the velocity of the air above the rotor disc? At some distance, it will sooner or later approach zero (this isn't apparent on your link, presumably because it's talking about a propeller that will be moving through still air).

In other words, I don't immediately see that the ratio can't be larger than 1:2, though I'm open to education and I'll admit I would have probably guesstimated the ratio to be less.
See my comments above- but also realize, Im assuming the velocity delta through the rotor is not decreasing with increased machine speed. That would need to be measured, and depends on the speed, but generally, it would drop (at the speed and angles in question, AOA should drop about 1deg inboard and about .4deg outboard, but math check would be appreciated).

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I would be very surprised if there are references where you can just look this up. Working it out homebrew, as you have, seems to me to be the only way (unless some kind hearted soul has posted their measurements, or unless perhaps you work for Align!).
I would be surprised also. No, I don’t work for Align, and I would suggest based on designs I have seen that most RC helicopter manufacturers do not employ aerodynamicists…
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Old 10-04-2011, 08:46 PM   #33
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I would be surprised also. No, I don’t work for Align, and I would suggest based on designs I have seen that most RC helicopter manufacturers do not employ aerodynamicists…
I've often thought that canopy design was a bit funny, but never really managed to work out how it should be optimised anyway - fast sideways flight; inverted; backwards...

Likewise, never sure whether going finless would actually make the helicopter aerodynamically neutral given that you also have a huge canopy up at the front and a tailrotor at the back... Haven't got the time or the skills to work it out so I'd be interested if you have any ideas.
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Old 10-05-2011, 03:14 AM   #34
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Right- define the optimization criteria. For most, it is ‘how cool does it look?’.

I think that you will find that, for all the marketing smoke and mirrors, this hobby is not very scientific. There is a lot of inertia in designs, and Id bet $1 that there is no comprehensive test data from any major blade supplier on their blades’ dynamic aero characteristics.

We can barely get base stats on our servos (i.e. show me official manufacturer specs on voltage vs torque vs angular rate). How about charts defining an ESC’s capabilities in various corner conditions (i.e. ability to react to a load transient), or its power efficiency into some standard motor across RPM/load.

In the absence of this data, it seems there are a whole lot of uncontrolled variables involved…
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