Item 1367
DESIGN: UniCopter - Rotor -
Blade - NACA00XX - Tip End Layout
To work up layout of the Torque Tube, Spar, Elastomeric bearing and Skin, etc.
Consider making the outer tip holder as part of the blade. Make out of out of carbon composite and include the tip end of the blade as part of it.
Drawing:
THIS COMPLETE DRAWING (with its extension to root) HAS NOW BEEN COPIED TO 1443 (OVERVIEW & LAYOUT) FOR DEVELOPMENT THERE OF COMPLETE BLADE.
THIS DRAWING IS PROBABLY NOW OBSOLETE.
Working Notes:
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Radius: |
r/R |
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Profile: |
Chord: |
Thickness at 25% |
Thickness at 30% |
Location of Spar on Chord |
Thickness at Spar C/L |
Inside Skin at Spar C/L |
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117.0" |
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Very tip |
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114.0" |
1.000 |
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NACA0006 |
7.0000" |
0.4158" |
0.4200" |
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0.4158" |
0.3118" |
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106.7" |
0.940 |
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NACA0007 |
7.8321" |
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0.5482" |
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NACA0008 |
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92.0" |
0.807 |
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NACA0009 |
9.5083" |
0.8500" |
0.8557" |
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0.8523" |
0.7483" |
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70.0" |
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NACA0012 |
12.0171" |
1.4324" |
1.4421" |
30 - (70/114) * 5 = 26.9298% |
1.4363" |
1.3323" |
On a NACA 00XX airfoil the pitch axis is at 25% of chord.
The thickest location is at 30% of chord.
Have the pitch axis at 25% of chord at the tip. Have the pitch axis at 30% of chord, or greater, at the root.
At NACA0012 the leading edge was 3.0039" ahead of the pitch line ( at 25% of chord). The airfoil at this location has been advanced further forward so that it is now 3.2362" ahead of the pitch line (at 26.9298% of chord).
Assuming that the cloth is 0.013" thick and there are 4 plies then the skin is .052" thick.
Currently on the drawing; the skin is 0.0607" thick at NACA0006, it is 0.0771" thick at NACA0009 and it is 0.0917 at NACA0012. It must be the same at all.
Torque Tube:
Thickness of tow 0.0279".
The span distance between the start of successive layers of torque tube winding is 1.8628". In other words, the taper of the airfoil at the location of the pitch axis is 2 * 0.0279" larger every 1.8628" of span.
For information on the layers see; DESIGN: UniCopter ~ Rotor - Blade - NACA 00xx - Torque Tube ~ TABLE: Torque Tube - Layer and the following notes.
Calculation Related to Laying out Tow on Drawing
- First Layer:
- Length of one tow along X-axis = Tow thickness / Wind Angle [Bias] = 0.0279" / Cos(54.9811) = 0.0485"
- Second Layer:
- xx
Third Layer:
xx
Fourteenth Layer:
Length of one tow along X-axis = Tow thickness / Wind Angle [Bias] = 0.0279" / Cos(64.2602) = 0.0642"
The following will all be modified.
Second layer: Not yet shown on drawing.
- The diameter in the center of the second wrap is 0.2713". The circumference is therefore 0.8527".
- Tan (Wind Angle) = 0.8527" / 1.8628" = 0.4577" therefore the wind angle is 25º.
- at 25º the distance of one tow along the span axis will be the tow thickness of 0.0279" divided by sin(25) = 0.0279" / 0.4226" = 0.0660"
- 0.0660" goes into 1.8628" 28 times. In other words, the bands have 14 tows.
Third layer:
The span distance between the start of successive layers of torque tube winding is 1.8628".
The diameter in the center of the third wrap is 0.3271". The circumference is therefore 1.0280".
Tan (Wind Angle) = 1.0280" / 1.8628" = 0.5519" therefore the wind angle is 29º.
at 29º the distance of one tow along the span axis will be the tow thickness of 0.0279" divided by sin(29) = 0.0279" / 0.4848" = 0.0575"
0.0575" goes into 1.8628" 32 times. In other words, the bands have 16 tows.
Fourteenth layer: (at NACA0009) Not yet shown on drawing.
- The span distance between the start of successive layers of torque tube winding is 1.8628".
- The diameter in the center of the fourteenth wrap is 0.8014". The circumference is therefore 2.5187".
- Tan (Wind Angle) = 2.5187" / 1.8628" = 1.3521" therefore the wind angle is 53º.
- at 53º the distance of one tow along the span axis will be the tow thickness of 0.0279" divided by sin(53) = 0.0279" / 0.7986" = 0.0349"
- 0.0349" goes into 1.8628" 53 times. In other words, the bands have 26 tows.
Fifteenth,sixteenth & seventeenth layer and higher:
- These will probably be single tow wound at a wind angle of ~ 90º
- 17 * 0.0279" = 0.4743" is the thickness. it is probably too much particularly considering that the diameter is increasing,
Notes:
The wedges will have the same width as the distance between quills because the tow thickness is the same in the torque tube and the spar. BUT the torque tube has a shallower taper.
If the tow is bent around a tight radius at the ends, it will become resin-starved. For this reason it might be advantageous to have the tension/return bar tool at the tip be outside (beyond) the finished product.
Consider an aluminum male and a pair of steel female mold halves. so that the higher expansion rate of the aluminum during cure will compress the tows.
Since oil is not being used to apply pressure, consider electrical wire heating.
The transmission of torque by the Torque Tube might be the weak aspect of the blade. Therefor consider having the tip end of the blade a lot of torque tube and little spar. In other words, have no spar over or under the actual torque tube near the tip. Have the only covering in this area be the skin.
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Last Revised: January 2, 2007