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Scorpion Power  Scorpion Power 
04122010 05:36 AM 8 years ago  Post 1 
alexz rrApprentice shakopee mn usa  I have a t600n with a 50 hyper.
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.  

04122010 07:17 AM 8 years ago  Post 2 
cudaboy_71 rrElite Veteran sacramento, ca, u.s.  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 600620s and a .90 can swing 690710s, it seems clear that more than 20mm difference in blade length changes the class of heli.if it ain't broke, break it.  

04122010 03:05 PM 8 years ago  Post 3 
QuantumPSI rrElite Veteran Atlanta, GA 
you'll lose about 5% of your lift  

04122010 05:26 PM 8 years ago  Post 4 
cudaboy_71 rrElite Veteran sacramento, ca, u.s.  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^2or 600mm blade area: 12.173695861154 sq/f 550mm blade area: 10.229286105553 sq/fPower Loading PL [hp/ft^2] = power / A or 600 PL : 0.164288645191 550 PL : 0.195517065352Thrust Loading (McCormick) TL [lb/hp]= 8.6859 * PL^(0.3107) or 600 TL: 15.223887284346 550 TL: 14.422606284722and, finallyLift = TL * power >>>[lb] = [lb/hp] * [hp] 600 Lift: 30.447774568691 lbs 550 Lift: 28.845212569444 lbs if it ain't broke, break it.  

04122010 05:35 PM 8 years ago  Post 5 
QuantumPSI rrElite Veteran Atlanta, GA  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 = BSdx/dt  

04122010 06:01 PM 8 years ago  Post 6 
cudaboy_71 rrElite Veteran sacramento, ca, u.s.  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.  

04122010 06:04 PM 8 years ago  Post 7 
QuantumPSI rrElite Veteran Atlanta, GA  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 = BSdx/dt  

04122010 06:30 PM 8 years ago  Post 8 
trekrider586 rrVeteran Sylacauga,Al 
 

04122010 06:55 PM 8 years ago  Post 9 
QuantumPSI  rrElite 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 = liftL = .5 * rho * v^2 * s * CLThis 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 * CLTo 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 bladesThe percent loss of lift is difference in lift between 600 and 550 blades/lift of 600 bladesOr 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 conservativeSo I would guess anywhere from 2326% 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 = BSdx/dt  

04122010 09:03 PM 8 years ago  Post 10 
cudaboy_71 rrElite Veteran sacramento, ca, u.s.  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.  

04122010 10:55 PM 8 years ago  Post 11 
QuantumPSI rrElite Veteran Atlanta, GA  Cudaboy
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  

04122010 11:06 PM 8 years ago  Post 12 
cudaboy_71 rrElite Veteran sacramento, ca, u.s.  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 rotorssimply 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.  

04122010 11:14 PM 8 years ago  Post 13 
QuantumPSI rrElite Veteran Atlanta, GA  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? Would the 550 sized blades lit the head speeds come back faster? how would this apply to an electric? and you could set the headspeed where you want,with a loss or gain in efficiency?  

04132010 01:29 AM 8 years ago  Post 14 
svenaslav rrNovice Kalamazoo, MI  

04132010 01:34 AM 8 years ago  Post 15 
helicoop rrApprentice North West Ohio  Was always told 90% of the lift was in the last 10% of the blade.  

04132010 04:36 AM 8 years ago  Post 16 
alexz rrApprentice shakopee mn usa  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.  

04132010 05:19 PM 8 years ago  Post 17 
QuantumPSI rrElite Veteran Atlanta, GA  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 = BSdx/dt  

04132010 05:35 PM 8 years ago  Post 18 
Rotowerkz rrApprentice Windham, NH  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 FXH 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.  

04132010 07:08 PM 8 years ago  Post 19 
Terrabit rrElite Veteran Seattle, WA  USA  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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