Shrink Wrap Wings !

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Grumpy Cynic
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Do you think a fabric covered airplane would have low-enough drag otherwise to benefit from active boundary layer control, and all the weight and complexity associated with such a system?
No! But I could be wrong too. There might be a remote chance it could be used to increase the CL max for a low stall speed?
 

Topaz

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Circulation control, perhaps, but I don't think boundary layer control significantly delays the onset of stall to a higher angle of attack. I don't have any information on that.
 

Topaz

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I knew I'd read this somewhere. Took me a few minutes to find it.

Pg 233/234 of TOWS. Not exactly the same thing. I've never seen any research on a porous surface but if one suction slot can get you to a CL of 3 it makes one wonder.
That would be "circulation control", rather than "boundary layer control." The former was a Big Thing in the 1960's and '70's, before someone pointed out that, if you depended on circulation control to land at your design-mission airports, what happens if the system breaks down and you're stuck with plain-old "lift"? Divert and hope there's a big-enough field within range? The other big application was in the 1980's when the NASA "X-wing" stopped-rotor demonstrator used it to allow reversed flow on half the rotor-wing once it was stopped in forward flight. Apparently that didn't work as well as they'd hoped.
 

airplaneAddict

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I thought of stretching shrink tape around each full rib to hold the fabric in place.

The shrink tape will stretch when pulled and snap back to it's original shape.
 

proppastie

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I thought of stretching shrink tape around each full rib to hold the fabric in place.

The shrink tape will stretch when pulled and snap back to it's original shape.

Not sure we can offer any more help, as you are in "uncharted territory", with the only other reference at post 23, and even that is probably a different stuff than you are using. One vendor advertised a filament tape for these type of plastics, but again, how it will stick, or hold up outside for even one hot sunny, wet rainy or cold snowy day is unknown, or a dive at 80 mph. Good luck please let us know ALL the results.
 

airplaneAddict

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the plane has been outside for almost a month...tape still holding well..survived sun, wind, rain, and hail..shrink wrap still tight...just mounted a BRS 750 chute to the plane, just in case...still have a few bugs to work out on the plane, and then I'll try to start the motor, and do a few hops down my dirt and grass runway
 

proppastie

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..just mounted a BRS 750 chute to the plane, just in case.
Good move. Remember a homebuilt will handle much different than C-150 trainer, see if you can get some dual in a 2 plc. ulta lite. I crashed on my first flight of a Midget Mustang, because I did not do that. Again good luck.
 

airplaneAddict

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I tried to find a dual trainer, but no luck...I'm practicing landings using a Drifter, and similar ultralights on MS Flight Simulator X...I do have rudder peddles...just like all simulators you can fly any plane that works perfectly...we both know an untrimmed plane won't behave that way...I'll wear a helmet, just in case.
 

proppastie

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General question to all the Stress guys: I wounder if this is applicable......:

Cable Loads
Tension in cable supporting uniformly load



cable load

The midspan force in a cable, wire or rope supporting a uniformly distributed load can be expressed as

H = w L2 / (8 d) (1)

where

H = midspan force in the cable (lb, N)

w = unit load (weight) on the cable (lb/ft, N/m)

L = cable span (ft, m)

d = cable sag (ft, m)

mass and weight

The force at the support ends of the cable can be estimated as

T = (H2 + (w L / 2)2)0.5 (2)

where

T = force at supports (lb, N)

The length of cable can be approximated to

S = L + 8 d2 / (3 L) (3)

where

S = cable length (ft, m)



kip = 1000 lb
klf = kip per linear foot

Example - Uniform Cable Load, Imperial units

A cable with length 100 ft and a sag 30 ft has a uniform load 850 lb/ft. The midspan force in the cable can be calculated as

H = (850 lb/ft) (100 ft)2 / (8 (30 ft))

= 35417 lb

The force at the supports can be calculated as

T = ((35417 lb)2 + ((850 lb/ft) (100 ft) / 2)2)0.5

= 55323 lb


http://www.engineeringtoolbox.com/cable-loads-d_1816.html
 
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airplaneAddict

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Boat Shrink Wrap and boat shrink tape covering holding up well outside. No peeling of tape, covering still very tight.

Got the motor started, but two of my tie downs ropes broke, and plane pivoted around the third on on the left wing.

On the way the prop managed to mangle a big step stool.

I'm excited the engine ran, but now I'm saving for a prop.

Need left hand pusher prop 68 x 30 or there about...got a prop for me?
 

airplaneAddict

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I think it was a drone...I don't know the specifics, but look how easy it was to apply if you have the $600 gun...I used 1500w heat gun and it worked fine, but slowly...still two weeks to cover the entire plane is fast
 

proppastie

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I keep thinking now is the time to get some scraps to play with at the boat yard
 

Midniteoyl

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shrink wrapped PoorBoy ultralight is now inside for the Indiana winter. 5" of snow for us yesterday.
Well shiver me timbers, I am 40mins North of you in Knox :)

Snow ticked me off yesterday.. I still have 5 more loads of wood to get and stack, but got put off due to 2 people having health problems that required my attention and I figures I had at least another week to get it done.. Nope, BOOM! Winter is here! :D
 
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