Item 0932
DESIGN:
UniCopter ~ Rotor - Blade - NACA 00xx - Spar ConstructionInformation and calculations related to the load carrying capability of the blade and its rigidity, etc, etc.
For general information on spar see;
DESIGN: UniCopter ~ Rotor - Blade - General - Spar![]()
Picture of Proposed Filament Winding:
Produce the complete spar at one time. The spar will be cured with a vertical split, so that there is a leading and a trailing half. The split will be widest at the tip and nonexistent at the root. At the tip it will only be wide enough to accept the tension/rotation tool.
This split will be opened slightly, after the cure, for the insertion of the torque tube, its elastomeric outer covering and the uncured epoxy overlay. It will then be closed, 90º bias wound and cured.
The tip end of the torque tube will be visible from the end?
Picture of half of mockup.
Think this through again.
What about the gap between the halves and the balder when the spar (only) is being pressure cured.

Leading or trailing half spar shown.
First layer shown. Most threads in each subsequent layer will wrap around a more distant quill.
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Variation in Changes in Tow Length Due to Active Blade Twist about the Feathering Axis:
Temporarily see
DESIGN: UniCopter ~ Blade - NACA 00xx - Core(s) for info on spar.
Blue is the end view of the tow when there is no twist
Red is the end view of the tow when the root has been twisted 15-degrees in respect to the tip.
At azimuth 0 & 180 the length of the tow does not change.
At azimuth 90 & 270 the tow stretches by 8.7% of the root's radius.
At azimuth 45, 135, 225 & 315 the tow contracts by 20.5% of the root's radius.
At azimuth, 82.5 & 277.5 the length of the tow changes during the rotation, but is equal at the start and finish.
This may result in an uneven loading of the centrifugal force on the individual tows and on the blade's attachment bolts.
Note that the change at 45º is greater than at 90º. The change at 45º will be slightly different than that which is shown on the drawing, since the spar at the tip will be thicker at 25% of chord then it is further forward and aft. In other words, the tows coming from 45º at the root will be closer to the feathering axis at the tip then is shown on the drawing.
It looks like the change in length at 45º will be Sine(0.3075 / 80) * 8 0.3075 = 0.0001"
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Compaction:
To compact the long tows would it be feasible to bind them by hoop wrapping for the full length of the span; perhaps using glass to give lest resistance to active twist?
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Winding Tension:
The tension on the tow should not exceed a pound or two, when it is being placed. This tension is probably also applicable to the tension along the span of the spar when closing the molds and curing. Perhaps if an elastomeric wrap was over the tip-puller and each layer then the pull will be distributed over many threads ???
Since the tow is not to twist as it is being laid and when being wrapped around the 'coned' pines at the root end the distance will be less on the strands nearer the tip of the 'Cone', can;
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Drawing:
Currently, this drawing is almost identical to that of 1054.html. This drawing will concentrate more on the spar.
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X-sectional Area of Spar
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Span: |
Section: |
Thickness: (1) |
Area: |
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10" |
3" OD. Ring |
3.0000" |
2.1607 sq-in |
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48" |
NACA 0015 |
" |
2.1607 sq-in |
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70" |
NACA 0012 |
" |
2.1607 sq-in |
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92" |
NACA 0009 |
0.7032" |
2.1607 sq-in (2) |
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114" |
NACA 0006 |
0.2986" |
0.8750 sq-in |
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117" |
~ |
0.0000" |
0.0000 sq-in |
(1) The thickness is that of the spar, not the airfoil. It is at the thickest point not at 25% of chord.
(2) Beyond roughly NACA 0009 the number of tow layers starts reducing. This is 25" from the very tip.
Get the quills much closer to the tip.
The taper of the spar in thickness is; ((3.0000 - 0.2986) / 2) / (114 - 10) = 0.01299" per Inch
= tan (0.01299) = 0.00022672º
If 12K tow with epoxy has a thickness of 0.0279" then the end of each layer of tow will be 0.0279 / 0.01299 = 2.1478" apart. In other words, the wedge will be 2.1478" wide.
If there is only one layer of spar at the tip, then there will be 25" / 2.1478" = 11.6 wedges.
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Width of Layers:
An ID of 2.5" = 7.8571" circumference, and 3.9286" half-circumference (width when flat near tip).
An OD of 3" = 9.4286" circumference, and 4.7143" half-circumference (width when flat near tip).
An OD of 2.875" = 9.0357" circumference, and 4.5179" half-circumference (width when flat near tip).
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Diagonal Winding at Root:
Place diagonal winding at the root, from the pins out toward the tip until reaching the location where the actual airfoil starts. The reason for this is to not 'waste' active twist in that portion of the span where it will do no good. This will reduce the size of the active twist and it will cause the airfoil root to foolow the root pitch horn exactly.
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A Thought on Strength:
The spar thickness may increase linearly from the tip to the root. If the area of the pultruded spar was not to change from tip to root then the out-of-plane strength will increase but the inplane strength will weaken. The inplane strength may be increased by having strong pultruded leading and trailing edges.
Consider doing one wrap at a wind angle of << 90º over the spar assembly. Use carbon tow from the tip in to the start of the elastomeric twist bearing then use glass for the rest of the way to the root. The glass should not give much resistance to the twisting.
A Thought on Curing the Spar:
Produce a latex sleeve twice as long as the span of the internal space inside the spar. This sleeve is rolled onto the torque tube form the root end up to the elastomeric 'bearing' (just like a long long condom). It is then rolled back down the inside of the spar. The assemblage is then placed inside an external female mold and hot oil is used to pressurize the 'condom'. The condom may remain inside the assemblage after curing or a release agent (Vaseline :) mat be applied to the inside of the spar etc.
After curing, the spar is partially opened at the root end an a couple of elastomeric bearing etc. are added.
To apply pressure on the tow that passes over the pin at the root, consider adding the 'plate', which will go between the blade root and the pitch horn, at this point and bond it to the spar.
A Thought on Thread Tension:
Smaller tow (say 12K or 6K) will have less problem with unequal thread lengths due to wrapping over a radiused surface.
Smaller tow (say 12K or 6K) may result in a better-finished root end. This is because the tow might be used to totally fill in the V's between the blade attachment bolts and therefor the bolts might be slightly shorter..
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Consider Producing the Spar in Eight Segments:

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Last Revised: September 24, 2005
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The stuff below here is from obsolete drawing 1067, review and ??
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Rejected Alternative - Spar with Wide Flange I-beam Profile at Mid-span:
Temporarily here.Am going with the oval profile as shown above, since twisting it results in significantly less differential length change in the tows.
Note that that at mid span the spar consists of 2 I-beams, This must be stronger in the out-of-plane direction than an
oval shape. This might be weaker than the oval shape in the in-plane direction.
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Calculations re carrying the same pultruded rods from the tip to the root: The following might go on 0886 and just be referenced here.
At tip NACA 0006 the chord is 6" therefore the maximum thickness, including skin is 0.36"
The maximum half thickness is 0.18".
Thickness of cloth (from SynchroLite VR-7) is 0.013" and there is two plys.
Maximum thickness of half spar is 0.154"
Area and strengths based on 27 pultruded rectangular rods:
Area of half spar is (13 * .02024) + ( 12 * .01009) + (2 * .00664) = 0.39748 sq-in.
Tensile strength = 320,000 psi. * 0.39784 = 127,309 pounds.
Compressive strength = 275,000 psi * 0.39784 = 109,406 pounds.
Theoretical centroids distance at :
8.8" chord @ theoretical location NACA 0027 gives a thickness of 2.376"
Less ((4 * skin thickness of 0.013) + (2 * pultruded rods thickness of 0.1490)) / 2 = 0.201
Centroids distance = 2.376 - 0.201 = 2.175" In actuality this is too great, but
Axial and Radial Forces and Lift:
From:
DESIGN: UniCopter ~ Rotor - Disk - Blade ForcesTorque Force = 2925 in-lb / 3 blades = 975 in-lb
Thrust Force = 8221 in-lb
Ultimate Radial load = square root (4,3572 + 49,3262) = 49,500 lb * (10G / 6SF) = 82500 lb
82,500 / 2.155 = 38,283 lb
Adjusted by centrifugal force: From:
DESIGN: UniCopter ~ Rotor - Disk - Blade ForcesUpper: 38,283 - 11,371 = 26,912 lb in compression
Lower: 38,283 + 68226 [ultimate centrifugal force] = 106,509 lb in tension
109,406 pounds / 26,912 lb = 4.06 SF
127,309 pounds / 106,509 lb =1.19 SF
The above is very crude and must be checked, but it looks like all will work
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