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F8F Bearcat - 125 MPH for $250

Old 02-09-2011, 03:45 PM
  #51  
NJSwede
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Originally Posted by z-8 View Post
It is simple aero that tapering-to-0 or elliptical wing tip (= 0 chord at the tip), twisted to neutral AoA (= 0 alpha at the tip), generates 0 lift at the very tip. That is exactly why props both twist and taper: Cl is not of the same degree and V^2, S is not of the same degree as V^2, as multiple terms taken together, they are.
Your math doesn't work out. I'm holding a 9x6 propeller for a HZ SuperCub in my hand, and from what I can see, the twist is pretty much linear along the length of a blade. Assuming the airfoil profile, area and air density are constant, we can write the lift equations as:

L = alpha * k * v^2 (assuming that lift is linear to alpha up to the stall point, which we can easily show that it is for the types of airfoils we're dealing with here). k is a constant that's dependent on the airfoil design, air density and all that jazz, which we can ignore for the sake of this discussion, since it's a constant.

Now, alpha decreases linearly as we move towards the tip while v increases linearly. But since L is proportional to v SQUARED, the increasing speed will have a much higher effect on the lift than the decreasing alpha. Thus, the lift will be higher toward the tip.

Looking at the propeller I have sitting on my desk, I also see that the width, and thereby there area decreases as you get closer to the hub. Since the lift is also proportional to the area of the airfoil, you're going to have even less thrust towards the middle.

Also, having all the thrust toward the hub would be counter intuitive. That's where you have the most obstruction from cowl, radial engine, air intakes and God knows what...
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Old 02-09-2011, 03:58 PM
  #52  
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Originally Posted by z-8 View Post
That is an old wives tale with no aerodynamic basis.
Then explain why helicopter rotors (which are optimised to hover rather than move through the air) have no twist???

The answer by the way is that twist is there to do exactly what I said.. To ensure each part of the blade operates at it's optimum angle of attack along it's full length. This will of course result in the outer part of the blade producing more thrust due to the fact that it's travelling faster through the air.

Here's a quote from the National Air and Space Museum web site:
Twisting the blade makes it meet the air at about the same angle across its entire length. This provides the most thrust and the least drag.
http://www.nasm.si.edu/exhibitions/g...tf/HTF531A.HTM

Of course propeller design is a complex business and twist can also be used to manipulate lift disribution just like you might do with twist on an airplane wing... But constant thrust (as opposed to constant coefficient of lift) along the span of the blade is certainly not what the goal is...
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Old 02-09-2011, 04:21 PM
  #53  
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Originally Posted by JetPlaneFlyer View Post
That's incorrect. The twist on the blade is to achieve constant pitch (and therefore approx constant blade angle of attack when in flight) not constant lift. The outer part of the blade will always lift MUCH more than the inner part due to it's velocity being higher, and lift being proportional to velocity squared.
In inner 1/3rd of the blade actually produces only a tiny proportion of overall thrust, probably 5% or less (which is why you don't lose any appreciable thrust when you fit a spinner). So losses due to the cowl will be vastly less than you might expect just based simply on occluded area.

Steve
Agreed, and simple to prove:
Set up motocalc with a motor that turns a propeller at some specific
RPM. Then check the predicted thrust. Then reduce that prop diameter to 1/3 the original diameter, and re-check the predicted thrust.

Example, a Hacker A40-12S with a 12X6 prop turns 10,300 RPM, producing 120 ounces of thrust. Put a 6 inch prop on it, readjust parameters to get the same 10,300 RPM, the predicted thrust drops to 7 ounces. (Did try a 4 inch diameter, 6 inch pitch prop, that would likely be stalled. But it did show 1.7 ounces thrust.) And, the watts input went up by a factor of 10 on the two different props.

Going up in diameter by an inch or two can cause a significant increase in watts input to your motor, and push it beyond its ratings. That's why it's so critical to have access to a wattmeter when trying different props on these motors.

I've got a model with an 12 inch diameter prop, 6 1/2 inch diameter cowl, it flys just fine. (Outside of the drag that cowl has on the model)
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Old 02-09-2011, 04:29 PM
  #54  
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Originally Posted by kyleservicetech View Post
I've got a model with an 12 inch diameter prop, 6 1/2 inch diameter cowl, it flys just fine. (Outside of the drag that cowl has on the model)
...and here's a model that's being praised for its flying characteristics all over this message board. I'd say about 40% of that prop is obstructed by the mock radial engine...

[media]http://www.parkzone.com/ProdInfo/Gallery/PKZ5080-GAL3.jpg[/media]
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Old 02-09-2011, 04:36 PM
  #55  
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Originally Posted by JetPlaneFlyer View Post
Then explain why helicopter rotors (which are optimised to hover rather than move through the air) have no twist???

The answer by the way is that twist is there to do exactly what I said.. To ensure each part of the blade operates at it's optimum angle of attack along it's full length. This will of course result in the outer part of the blade producing more thrust due to the fact that it's travelling faster through the air.

Here's a quote from the National Air and Space Museum web site:

http://www.nasm.si.edu/exhibitions/g...tf/HTF531A.HTM

Of course propeller design is a complex business and twist can also be used to manipulate lift disribution just like you might do with twist on an airplane wing... But constant thrust (as opposed to constant coefficient of lift) along the span of the blade is certainly not what the goal is...
That's what I said, twist toward the tip reduces the effect of increasing radial velocity toward the tip. Both airfoil twist and a reduction in chord, taken together, are designed to reduce lift as a function of radius to avoid breaking the cantilever--meaning both taper (S) and twist (Cl), multiplied together in the L equation, combine to offset an otherwise increasing L from radial velocity, where the effect is proportional to V squared.

Obviously, there are different designs and constraints with helicopters and some constant/fixed speed prop designs, but most simple, light, and reasonably efficient props both twist and taper.
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Old 02-09-2011, 04:38 PM
  #56  
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Originally Posted by NJSwede View Post
...and here's a model that's being praised for its flying characteristics all over this message board. I'd say about 40% of that prop is obstructed by the mock radial engine...
Largely irrelevant. The bulk of the thrust comes from the end of the prop. Due in large part to the fact that the tips are spinning the fastest.
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Old 02-09-2011, 04:39 PM
  #57  
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[QUOTE=NJSwede;784012]...and here's a model that's being praised for its flying characteristics all over this message board. I'd say about 40% of that prop is obstructed by the mock radial engine...

QUOTE]

Assume that prop is 12 inches diameter, and the cowl is 6 inches diameter. So the propeller disc has an area of Pi*R^2, or 113 square inches. And the cowl has an area of 28 square inches. That alone shows that the "active" area of the prop is four times the cowl area.
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Old 02-09-2011, 04:42 PM
  #58  
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Originally Posted by rcers View Post
Largely irrelevant. The bulk of the thrust comes from the end of the prop. Due in large part to the fact that the tips are spinning the fastest.
That was sort of my point...

L = alpha * k * v ^ 2 tells the whole story. You'll have most of the thrust from the end of the blades since it's proportional to v-squared. Z-8 seemed to imply it didn't.
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Old 02-09-2011, 04:43 PM
  #59  
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Originally Posted by NJSwede View Post
...and here's a model that's being praised for its flying characteristics all over this message board. I'd say about 40% of that prop is obstructed by the mock radial engine...

[media]http://www.parkzone.com/ProdInfo/Gallery/PKZ5080-GAL3.jpg[/media]
Yep, flying tree trunks of all varieties are great examples of why installation error must be considered when propping. I actually use the Trojan as the chosen example in several places in my blog article on power system selection. Note the Trojan pic, the caption, and anecdote in the text.
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Old 02-09-2011, 04:46 PM
  #60  
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Originally Posted by rcers View Post
Largely irrelevant. The bulk of the thrust comes from the end of the prop. Due in large part to the fact that the tips are spinning the fastest.
There is more than just V^2 in the lift equation. Decreasing chord (S) and Cl, are painstakingly designed to offset the effect of V^2 in any prop that twists and tapers.
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Old 02-09-2011, 04:52 PM
  #61  
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Originally Posted by NJSwede View Post
That was sort of my point...

L = alpha * k * v ^ 2 tells the whole story. You'll have most of the thrust from the end of the blades since it's proportional to v-squared. Z-8 seemed to imply it didn't.
Here is the L equation:



A = surface area which is interchangeable nomenclature with S and a function of chord, or taper. Cl is roughly proportional to AoA which is why there is twist. The terms Cl * S are designed to offset V^2 in prop blades that twist and taper (and some aircraft wings too, where V is in that case constant, but the length of the cantilever would still cause it to break or be too heavy otherwise).
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Old 02-09-2011, 05:04 PM
  #62  
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Originally Posted by z-8 View Post
Here is the L equation:



A = surface area which is interchangeable nomenclature with S and a function of chord, or taper. Cl is roughly proportional to AoA which is why there is twist. The terms Cl * S are designed to offset V^2 in prop blades that twist and taper (and some aircraft wings too).
I am very well aware how the math works. However, if you hold the area, air density and CL constant, you get the formulat I gave your. CL can be broken down into angle of attack and a constant that's dependent on the airfoil design. This works out to

L = k * alpha * v ^ 2, where k is a constant dependent on area, air density and airfoil design.

Now, we can of course add the area back to the equation and get

L = A* k * alpha * v ^ 2

So you have pointed out that the chord length varies. Yes it does. It becomes SHORTER the closer you get to the hub. Since the chord length decreases, so does the area. In other words, if you cut off a millimeter of prop toward the hub, it will have a smaller area than a millimeter of prop toward the top.

Thus, the chord length will NOT cancel out any effect of the higher velocity toward the ends. It will AMPLIFY that effect. Can you please tell me how your math would work out?

Anyone on Wattflyer knows I'm a pretty lousy pilot and that I've never built a balsa model. But I *DO* know my math pretty well and I can't get your reasoning to line up with the equations I know...

If I am wrong, I'd love to be educated, so please explain how it works!
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Old 02-09-2011, 07:09 PM
  #63  
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Originally Posted by z-8 View Post
There is more than just V^2 in the lift equation. Decreasing chord (S) and Cl, are painstakingly designed to offset the effect of V^2 in any prop that twists and tapers.

Looks like Master Airscrew got it horribly wrong when they made their props then.

I just checked a 13x6 Master Airscrew prop.

I took measurements at two stations:
  • 1/3rd 'span' (55mm from hub): Blade width = 29mm, blade angle= 22deg.
  • Tip (165mm from hub): Blade width = 17.5mm, blade angle= 12deg.
Taper and twist is virtually linear between these points. Inboard of the 1/3rd 'span' the blade gets narrower.

The angle of the 'relative wind' that approaches the blade depends on tangential velocity of the blade and forward speed of the prop through the air. Simple trig shows the angle of the relative wind to be = arctan (forward velocity/tangential velocity)

Throwing in some typical data:

RPM = 10k
Forward speed = 60kmh (16.7m/s)
  • At the 1/3 blade station relative wind angle = arctan (16.7/ 0.11 x pi x 10,000 /60)) = 16.2 deg
  • At the tip = arctan (16.7/ 0.330 x pi x 10,000 /60)) = 5.5 deg

The angle of attack that the prop blade airfoil 'sees' is = blade angle minus relative wind angle

So
  • At 1/3rd blade span = 22 - 16.2 = 5.8deg AoA
  • At the tip = 12 - 5.5 = 6.5deg AoA

So what this proves is at a flying speed of 60kmh the tip is actually at a very slightly higher AoA than the 1/3rd blade span station. Within the accuracy of my blade angle measurement we might as well call it equal.


Going back to the lift formula:
for comparison purposes we can neglect air density as it's constant. This means lift that the blade produces along it's span is proportional to only velocity squared, angle of attack and area.

The tip is travelling 3 times faster than the 1/3rd span stationt.. 3 squared is 9, so if all else is equal the tip would make 9x more lift.

Angle of attack has been shown to be the same so that can be neglected in our comparison.

Blade area is less at the tip, in proportion of 17.5/29 = 0.6

Boiling it all down it turns out that the lift (hence thrust) produced by the tip is 9 x 0.6 = 5.4 times greater than the thrust produced at the 1/3rd station. (****see edit****)

Comments welcome

Steve

PS.. The above calculation demonstrates why helicopters don't use twisted blades. This is because their forward speed (or 'upward speed for a helicopter) is optimised for the hover so assumed to be zero. This produces zero relative wind angle and hence no requirement for twist.

PPS.. this does neglect tip losses due to tip vortexes but that is outside the scope of this comparison.

***edit*** I re-checked the tip angle and it's actually closer to 10 deg than to 12. This makes a minor difference to the calc, the tip lifts about 4.5 times more (not 5.4). This doesn't change the basic conclusion.

Last edited by JetPlaneFlyer; 02-09-2011 at 11:20 PM.
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Old 02-09-2011, 08:14 PM
  #64  
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well,

i might be way out of my league on this one as far as the math.

you guys no your math way better then i do but from the readings i have read and understand. that the angle of atack or pitch of the prop is the main facter in producing lift or thrust. constant lift or thrust is needed or the prop will stall in differant radiuses of the prop. so this is why there is a twist almost linearly in the prop.

for any airfoil to be efficient i do know it must have enough cord length to creat lift at a given speed and angel of attack or the airfoil will stall. there must be enough cord length there to function not realitive to area because you can have a real long wing with to short of cord and if the wing dose not hit the speed it needs to for the given cord length at a given angle of attack it will never lift off the ground or be efficient no matter what wing area you have. actually by lengthening the wing with to short of cord length it induces more drag thus making it harder to hit the opitmal speed of the cord length.

now the opposite is also true to long of cord length for a given angle of attack and to high of speed induces drag.

so you see you need just enough cord length to start lifting at a given speed and not to much cord lenght to induce to much drag no less no more to be realy efficient.

since the prop speed increases from the root to the tip so dose the drag on a constant cord prop. wing cord is lessened in the fact that the airfoil at such higher speeds is inducing to much drag. the wing cord realy is made less at the tip to decrease drag. not decrease the lift.

so on a twisted angle of atack prop the shorter cord is at the tip because the speed is much higher thus taking a shorter cord to produce efficient lift.


so this is why when you cut off the end of an apc prop it is not as efficient. and such that i believe the apc to be superior to a master airscrew prop.

this is just my reasoning and i can not prove it with math. oh and i defintaly could be wrong on this.. lol

Last edited by 57sailplane; 02-09-2011 at 08:50 PM.
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Old 02-09-2011, 08:21 PM
  #65  
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Originally Posted by JetPlaneFlyer View Post
Comments welcome
My only comment is that you missed the opportunity to take it even a step further. You're just looking at the lift at a point on the blade. Let's have fun and look at the lift along the entire inner half and the entire outer half. We then need to calculate the integral of lift along the two halves and take the ratio (which, of course, gets rid of that pesky "k"):



So, if my calculations are correct, we'd have 7 times more thrust coming from the outer half of the blade.

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Old 02-09-2011, 08:53 PM
  #66  
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All of these calculators are fine and dandy, but in the end they're just a guide to finding the right power system for a given model. As has been stated before, the same power system can be put in different models and have them produce radically different results.

For example, I'm a big fan of pusher jets (hence the webstore ). Two of my favorite planes are the profile F-22 and the profile Saab Draken. Both models weigh almost the same (within tenths of an ounce), and both are in the same "size category".

However, with the exact same power setup in both (motor/esc/prop/lipo all matched), the F-22 does about 55mph flat and level, while the Draken does closer to 75! This has been verified with radar.

YMMV
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Old 02-09-2011, 08:54 PM
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57sailplane, the easiest way to think about it is like this:

Think of a point in time and a point on a propeller blade. At any given time, that point will be subject to two forces: The resistance of carving through the air along its rotational path and the air rushing against it because of the movement of the plane. These two forces are perpendicular, i.e. have a 90 degree angle between them. To understand the actual air flow against the propeller, we need to combine the two forces. Mathematically, we call this a "vector addition".

Instead of going into the mathematics (I think we've had enough of that for today), we can think of it like this: Imagine you're in a convertible car with a 20mph crosswind. When the car is stationary, it's going to feel like the wind is coming from the side. However, if you accelerate the car to, say, 60mph, you're going to feel mostly headwind. This is because the speed of the car turns the "apparent wind vector" towards coming from straight ahead of you the faster you go.

Now think about the propeller again. A point near the hub turns very slowly compared to a point toward the tip. That means that the point towards the hub is going to experience mostly headwind (due to the movement of the plane), whereas the point close to the tip is going to experience mostly its movement along its circular path. Therefore, it makes sense to have a propeller that's twisted so that the areas close to the hub have their leading edge pointed more towards the direction of the headwind, since that's the airflow they'll mainly experience. The opposite is obviously true for the areas close to the tip.

Edit: The example with the car got a little confusing. For it to make sense, you have to think of the "headwind" of the propeller as the air it encounters along its rotational path and the "crosswind" as the wind generated by the movement of the car. Then in makes perfect sense! The point is that an airflow is going to "feel" different dependent on your speed relative to it.

Last edited by NJSwede; 02-09-2011 at 09:18 PM.
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Old 02-09-2011, 09:04 PM
  #68  
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Originally Posted by crxmanpat View Post
However, with the exact same power setup in both (motor/esc/prop/lipo all matched), the F-22 does about 55mph flat and level, while the Draken does closer to 75! This has been verified with radar.
That's easy to explain: Draken is a Swedish plane!

No, but seriously, the fact that reality isn't 100% clear cut and doesn't always match up with theory is one of the reasons I'm into this. If everything was completely predictable, it'd be pretty boring.

I think the point most folks here are trying to make is that math and calculators give you a rough but valuable idea of what reality might look like. It takes a lot of the guesswork out of things.
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Old 02-09-2011, 09:19 PM
  #69  
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Here is an efficient scimitar blade:

AOA at tip = less than 3 degrees = slight negative AoA at airspeed (most props push backward near the tips beyond the optimum/design airspeed)
Chord at tip = 0

Either one or both terms drive Lift to zero at the tip.
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Old 02-09-2011, 09:25 PM
  #70  
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Originally Posted by crxmanpat View Post
All of these calculators are fine and dandy, but in the end they're just a guide to finding the right power system for a given model. As has been stated before, the same power system can be put in different models and have them produce radically different results.

For example, I'm a big fan of pusher jets (hence the webstore ). Two of my favorite planes are the profile F-22 and the profile Saab Draken. Both models weigh almost the same (within tenths of an ounce), and both are in the same "size category".

However, with the exact same power setup in both (motor/esc/prop/lipo all matched), the F-22 does about 55mph flat and level, while the Draken does closer to 75! This has been verified with radar.

YMMV
Exactly. (I'm surprised they are that close)
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Old 02-09-2011, 09:37 PM
  #71  
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Originally Posted by z-8 View Post
Here is an efficient scimitar blade:

AOA at tip = less than 3 degrees = slight negative AoA at airspeed (most props push backward near the tips beyond the design airspeed)
Chord at tip = 0

Either one or both terms drive Lift to zero at the tip.
Ok. I give up. I've been trying to understand how you can get a negative angle of attack. If you look at the extreme cases, you won't even get a negative angle of attack from them:

Zero degree pitch, all headwind no rotation: 90 degree angle of attack.

Zero degree pitch, no headwind, all rotation: 0 degree angle of attack.

What am I missing here?
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Old 02-09-2011, 09:45 PM
  #72  
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Originally Posted by NJSwede View Post
Ok. I give up. I've been trying to understand how you can get a negative angle of attack. If you look at the extreme cases, you won't even get a negative angle of attack from them:

Zero degree pitch, all headwind no rotation: 90 degree angle of attack.

Zero degree pitch, no headwind, all rotation: 0 degree angle of attack.

What am I missing here?
Sure - imagine a prop that twists to flat (zero AoA at the tip). So at the tip, the AoA will be negative with any forward airspeed vector. The faster the airspeed, the more negative it goes. Most props don't quite twist to zero AoA, but at or beyond their design speed and RPM, they will zero out at the tip or go slightly negative Lift.
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Old 02-09-2011, 09:47 PM
  #73  
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And by the way, mathematically speaking, the chord at the end of the propeller isn't zero. It approaches zero, just as any propeller with a pointed or rounded edge does. If the chord is zero, you're by definition not ON the propeller but OUTSIDE of it.

But one thing is true in what you're saying. You *CAN* have "negative thrust" at the tip. Most notably, it happens when you deadstick. Since the area close to the tip is facing 90 degrees against the wind, it's generating a ton of drag. It's actually generating a ton of drag when the prop is turning too. The point is that since the rotational speed is so high and the lift increases by v-squared, we can afford the extra drag, since we're generating more "lift" (i.e. thrust) than the drag generated along the axis of the plane's motion. And THAT, my friends, is another way of explaining why props are twisted.

But I don't feel this discussion is going to be meaningful much longer, so I respectfully bow out and enter lurking mode!
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Old 02-09-2011, 09:49 PM
  #74  
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Originally Posted by z-8 View Post
Sure - imagine a prop that twists to flat (zero AoA at the tip). So at the tip, the AoA will be negative with any forward airspeed vector. The faster the airspeed, the more negative it goes. Most props don't quite twist to zero AoA, but at or beyond their design speed and RPM, they will zero out at the tip or go slightly negative Lift.
You're just restating what you just said. Not how it is possible. A diagram or calculation would be nice. But as I said, we have passed the limit of what is a meaningful discussion a while ago. Have a nice evening!
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Old 02-09-2011, 10:26 PM
  #75  
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Originally Posted by z-8 View Post
Sure - imagine a prop that twists to flat (zero AoA at the tip). So at the tip, the AoA will be negative with any forward airspeed vector. The faster the airspeed, the more negative it goes. Most props don't quite twist to zero AoA, but at or beyond their design speed and RPM, they will zero out at the tip or go slightly negative Lift.
Most (if not all) props will go to zero AoA at the root first. The tip will generally be the last part of the prop to stop producing thrust and so unless you were in a very steep dive it would not actually occur.

Taking my 13x6 Master Airscrew prop as an example. The 1/3 span station would at 10k RPM go to zero AoA at an airspeed of 83.63kmh (tan22 deg x tangential velocity)

The tip would still be operating at a positive AoA of 2 or 3deg at that airspeed.

Simple trig.. Calculus makes my head hurt but hats off to NJSwede for showing us how it's done. He appears to have a mathematical mind similar to the guy I sat next to and cribbed off to get me through my math exams

Steve

Last edited by JetPlaneFlyer; 02-09-2011 at 11:11 PM.
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