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HomeAircraftHelicopterMain Discussion › Question on what size blades?
04-12-2010 05:36 AM  8 years agoPost 1
alexz

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shakopee mn usa

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I have a t600n with a 50 hyper.
Can I use 550 main blades on a 50/600 size ship? I never thought to ask. I always thought that the 550 sized blades were for the 500/30 sized ships. But if you have a governor than I don't see why one couldn't. Come to think of it they are cheaper than 600 sized blades.

My flying stile is geared more to 3D flying. I wouldn't say hard 3D. I can do up right/inverted funnels, four point tick tocks, piro flips,and tail slides, along with some other mild tricks.

Would the 550 sized blades lit the head speeds come back faster? For some one like me that is still not that good at collective management.

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04-12-2010 07:17 AM  8 years agoPost 2
cudaboy_71

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sacramento, ca, u.s.

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i'd be reluctant to say 'No'.

but, to get the discussion rolling i'd offer that you'll lose about 5% of your lift right out of the gate, due to shorter blades. since nitro motors are tuned to run at a given RPM, you cant just change your headspeed to make up for the loss of lift. regearing *could* be an option. not having ever done it i'm not sure whether you'd be chasing the power band or not....

but, consider this: the difference between a .50 class heli and a .90 class heli is only a difference in blade length of 90mm. shortening a 600 to 550 is 50mm. factoring in that a .50 class heli can swing 600-620s and a .90 can swing 690-710s, it seems clear that more than 20mm difference in blade length changes the class of heli.

if it ain't broke, break it.

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04-12-2010 03:05 PM  8 years agoPost 3
QuantumPSI

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you'll lose about 5% of your lift
You'll lose more than 5%, most lift is generated at the tip (lift goes with v^2). I don't have time to work out the math, but I'm very sure you'll lose more than 5%. Running the same RPM, I'm sure the difference is significant.

...now where was I, dh/dt = BS-dx/dt
I will fly you forever... till earth do us part

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04-12-2010 05:26 PM  8 years agoPost 4
cudaboy_71

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i just quickly calculated the lift of a 1200mm and 1100mm disk (yes, i neglected the head diameter, losses to tail rotor, and losses to friction, etc... but for the sake of comparison it should yield the same ratio), and the difference in lift was ~30.448 lbs vs. ~28.845 lbs, or ~5%

if i used the wrong formulas or assumptions, then feel free to correct. but, assuming 2hp for a hyper 50, i used:

Area [ft^2] = Pi * r^2
or
600mm blade area: 12.173695861154 sq/f
550mm blade area: 10.229286105553 sq/f

Power Loading
PL [hp/ft^2] = power / A
or
600 PL : 0.164288645191
550 PL : 0.195517065352

Thrust Loading (McCormick)
TL [lb/hp]= 8.6859 * PL^(-0.3107)
or
600 TL: 15.223887284346
550 TL: 14.422606284722

and, finally

Lift = TL * power >>>[lb] = [lb/hp] * [hp]
600 Lift: 30.447774568691 lbs
550 Lift: 28.845212569444 lbs

if it ain't broke, break it.

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04-12-2010 05:35 PM  8 years agoPost 5
QuantumPSI

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Cudaboy, though your logic is sound, it's not quite right. Your calculation assumes the lift generated per sq. in. of rotordisk area is constant (or rather, the same everywhere on the disk). This isn't right because the tips are moving faster than the roots and thus the tips generate FAR MORE lift than the root. The lift distribution is parabolic along the length of the blade and not a constant value.

What you've done would be fine for a wing (where the air velocity is the same everywhere and neglecting tip effects), but since the velocities are different on a helicopter blade, the lift is far different along the length.

And after looking at your formula more closely, the problem arises when you calculate the power/area. The power requirement is not the same throughout the area of the disk (hence what I was saying up above). Tips are moving faster, and thus the drag is higher. Power is angular velocity*torque (looking at one slice of the blade, the torque varies with r^2). The reason the torque varies with r^2 is because it's moving faster and drag goes with velocity^2. The velocity a slice of the blade sees varies linearly with r, thus, drag goes with r^2.

Sorry for being long winded, I'll try to work out the numbers tonight. To do it accurately, you'll need to set up an integral.

...now where was I, dh/dt = BS-dx/dt
I will fly you forever... till earth do us part

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04-12-2010 06:01 PM  8 years agoPost 6
cudaboy_71

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well, to be honest my original thought was to just show how much lift would be lost by calculating the lift from the outer 50mm of a 600mm blade. but, i couldnt find a formula, and my calculus is 20 years too rusty to run the number on my own.

i still don't see the error in what i did post...the total lift generated by a disk is still a single number, regardless of whether most is generated at the tip or not.

but, if you can find the time to post your calculations i'd love to see them, if for no other reason than to be able to do it myself in the future.

thx.

if it ain't broke, break it.

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04-12-2010 06:04 PM  8 years agoPost 7
QuantumPSI

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Sure, no problem. I'll try to make a sketch as well to show what I mean. I think a picture would help immensely in this. (back to studying (this is getting old ))

...now where was I, dh/dt = BS-dx/dt
I will fly you forever... till earth do us part

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04-12-2010 06:30 PM  8 years agoPost 8
trekrider586

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Sylacauga,Al

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interesting
how would this apply to an electric?
say 600e.it's already a light machine,so weight should'nt be a factor.
and you could set the headspeed where you want,with a loss or gain in efficiency?

it's better to ask for forgiveness than for permission

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04-12-2010 06:55 PM  8 years agoPost 9
QuantumPSIrrElite Veteran - Atlanta, GA - My Posts: All  Forum  Topic

Ok so I made a crude sketch for what I mean. The blue part is the blade and the red part is the lift distribution. The total area of the red is the total lift that is generated. (A is the root, B is the tip) What I'm trying to say is that, because the distribution is not constant (straight across), removing a portion of the blade at the tip is the not the same as removing that same amount at say the root. You can see this in the picture. Assuming the lengths of section A and section B are the same, you can easily see that removing B will remove far more lift than A. This is the simple way to see it. Now for the math...

rho = density
c = chord
s = area of lifting surface
r = radius
v = velocity
CL = coefficient of lift
w = angular velocity
L = lift

L = .5 * rho * v^2 * s * CL

This equation applies to a wing where the velocity is constant throughout the length of the wing. But since we are dealing with helicopters and thus rotations, we need to convert this equation.

Area of the blade (just a slice)
ds = c*dr (assuming constant chord blades, which isn't a bad assumption) (function of r)

Velocity at each point on blade
v(r) = r*w (function of r)

Plugging these two back into the lift equation, we get
L = .5 * rho * r^2 * w^2 * c * dr * CL

To find find the total lift, we integrate this over the length of the blade (ie with respect to r).

Since we are just looking at the loss of lift, we don't need to know rho, c, w, or CL. The reason for this is that we are assuming that the helicopter is flying in the same air (not a bad assumption), spinning blades are the same RPM (again, not a bad assumption), have the same chord (ok, this isn't exactly true, the 550 blades will have a smaller chord, but this just means that actual percentage difference in lift will be HIGHER (ie, you will lose even more lift than I calculated)), and the same CL (same as the chord, not exactly true, but the 550 will have a lower CL due to smaller chord, so even more lift is lost than calculated).

For simplicity
L(600) = lift from 600 blades
L(550) = lift from 550 blades

The percent loss of lift is
difference in lift between 600 and 550 blades/lift of 600 blades

Or better yet
[L(600) - L(500)]/L(600) (call this eqn 'ratio' )

I'm assuming the root of the blades start at about 8cm from the mainshaft.

L(600) = .5 * rho * c * w^2 * CL * integral(r^2 dr)[68 8] (integral of r^2 from 8 to 68)

L(500) = .5 * rho * c * w^2 * CL * integral(r^2 dr)[63 8] (integral from r^2 from 8 to 63)

Plugging this in to 'ratio' and cancelling out terms (the .5, rho, c, w^2, and CL) we get

{integral(r^2 dr)[68 8] - integral(r^2 dr)[63 8]}/integral(r^2 dr)[68 8]

The integral of r^2 dr is (r^3)/3.

Plugging in all the numbers, and the difference you'll get is 21%. A HUGE DIFFERENCE! And mind you, the assumptions I made (same chord and same CL) make this a CONSERVATIVE estimate, meaning, the actual difference in loss of lift is higher than 21%. So removing those last 5 cm of the blade removes about 20% of the total lift of the blade. Kind of hard to believe at first, but when you realize that most of the lift is generated at the tips (see picture above), it makes sense. Mind, this doesn't account for tip losses either, so it's even more conservative

So I would guess anywhere from 23-26% loss of lift by using the shorter blades.

Edit: I made a mistake in one of the equations, sorry. It's fixed now

...now where was I, dh/dt = BS-dx/dt
I will fly you forever... till earth do us part

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04-12-2010 09:03 PM  8 years agoPost 10
cudaboy_71

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thanks for that. copied and pasted to my notepad for later ingestion.

now, back to my original confusion:

i fully understand and was fully aware that the outermost portion of a rotor blade generates more lift due to greater speed. my confusion was (still is) that total lift generated for a given disc area is a fixed number. sure, it may generate more or less at different parts of the disc. but, the disc as a whole generates a single value.

i still don't see how subtracting the total lift of a larger disc from the total lift from a smaller disc yields anything less than the difference in lift between the two....UNLESS....

the formula for lift of a disc I used is wrong...every reference i found online (which very well could be referencing the same bad seed of info) supported the equations i used above.

if it ain't broke, break it.

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04-12-2010 10:55 PM  8 years agoPost 11
QuantumPSI

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Atlanta, GA

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Cudaboy
Ok so I did some searching and found where you got your information from. From what I can gather, the information is based on a propeller. This is where the problem lies. I don't know if you've flown RC airplanes, but if you look at a propeller up close, you'll notice a few key features. First off, the airfoil changes along the length, as well as the angle of attack, and the chord. The whole point of this design is so that the propeller generates the same thrust along the length of each blade (hence the really high angle of attack at the root and really shallow attack at the tip). However, when you compare this our helicopters, it is very different. Everything is constant along the length (angle of attack, chord length (except for tapered blades), airfoil, etc.). For a WELL DESIGNED propeller, the lift distribution is roughly constant and what you've done up above would be correct. But as stated before, our helicopters aren't quite like that.

Another way to think about it. Imagine you have a plate as big as rotordisks and you're pouring sand over it (the sand represents lift). If you spread the sand over the disk evenly, then what you've done is correct and the lift generated is directly related to the area. However, with our helicopters, you would have to put more sand at the edge of the blade than at the center. And when you do this and remove a certain amount of area, where you remove that area will determine how much sand you remove. In this case, since you're removing a ring at the tip, you're removing the highest concentration of sand.

So yes, the total lift is a single number, but how/where that lift is distributed plays a CRUCIAL role when you're removing area from the lifting surface.

I hope this helps, but let me know if I've confused you anymore

...now where was I, dh/dt = BS-dx/dt
I will fly you forever... till earth do us part

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04-12-2010 11:06 PM  8 years agoPost 12
cudaboy_71

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thanks for clearing that up. i think it all boils down to i used the wrong formula. at least one place online (can't find the link now) used those same formulas for both propellers and heli rotors--simply replacing the word 'thrust' with 'lift' in the various examples of use. that one bit of bad info led me down the wrong path.

if it ain't broke, break it.

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04-12-2010 11:14 PM  8 years agoPost 13
QuantumPSI

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Just realized that the OP's questions haven't been directly answered, sorry.
Can I use 550 main blades on a 50/600 size ship?
If they'll fit, yeah you can, but the performance would probably suck unless you cranked up the HS but as cudaboy said, engines are designed for specific RPMs.
Would the 550 sized blades lit the head speeds come back faster?
Yes, the HS would recover faster, but other things would suffer as stated previously.
how would this apply to an electric?
This applies to all helicopters.
and you could set the headspeed where you want,with a loss or gain in efficiency?
This is kind of a vague question, but I'll do my best to answer anyway. It really depends. Electric helicopters are different from nitros. You could regear it (different pinion) to keep the motor at the same RPM and change the HS accordingly. Whether or not it would be more efficient is dependent on a lot of factors that are not accounted for in my little calculation. Off the top of my head, I can't say for sure if you'd gain or lose efficiency by changing blades to something smaller and increasing the HS. It really depends on a lot of things...

...now where was I, dh/dt = BS-dx/dt
I will fly you forever... till earth do us part

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04-13-2010 01:29 AM  8 years agoPost 14
svenaslav

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Kalamazoo, MI

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04-13-2010 01:34 AM  8 years agoPost 15
helicoop

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North West Ohio

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Was always told 90% of the lift was in the last 10% of the blade.

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04-13-2010 04:36 AM  8 years agoPost 16
alexz

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shakopee mn usa

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Lots of info there to chew on. Thanks you for all the work you put into given us those numbers.

Thank you and know I now and nowing is half the battle.

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04-13-2010 05:19 PM  8 years agoPost 17
QuantumPSI

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Just FYI, I was thinking about this while I was not paying attention in class today and I realized that my original formulation is incorrect, sorry . I've since corrected it and recalculated the numbers. The actually percentage loss is about 21% (not 26%). I had double counted an 'r' which should have been a 'dr' (calculus mistake, I won't explain further unless asked via PM, but it's mathematical in nature). In any case, it's been corrected now and again, sorry for the mistake.

With that being said, the loss of lift is still quite significant (just not as bad as I had calculated previously ) but still, bad nonetheless.

...now where was I, dh/dt = BS-dx/dt
I will fly you forever... till earth do us part

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04-13-2010 05:35 PM  8 years agoPost 18
Rotowerkz

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Windham, NH

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If you regear to compensate for the lower load on the drivetrain, and to keep from overspeeding the engine, you might be okay. The load on the engine is lower, so you would be able to get away with it if you used an aggressive governor setting. If you understand where the Hyper's torque band peaks out, and keep the RPM below that, you have a basis for keeping the engine happy.

I have a Raptor 30 with an OS .46 FX-H in it right now, and I really notice the reduced pop of the heli (when compared to a bird swinging 600mm blades). The engine is not real happy with the gearing. However, the .46 develops its torque at a lower RPM than the Hyper 50; it does fly better than the .32 that was in there previously.

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04-13-2010 07:08 PM  8 years agoPost 19
Terrabit

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The difference between 690's and 710's on my vbar t700 is enormous. Initially, the 91hz along with the flybar and my poor collective management couldn't really handle 710's without frequent bogging. It certainly rocks now though. Zero bogging!!!

I couldn't imagine dropping 50mm of blade length.

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