Minimax to Parasol

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BobDaly

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Posting this for your critiques. I'm converting my Minimax to a parasol configuration. Partly because mounting the wings was difficult and the bolt holes thru the axle for the lift struts were becoming elongated. I'm also partial to the parasol look. I figured a parasol conversion would work because the Himax is a proven design and I would merely be eliminating the center section and enclosed cockpit. I'd like some input as to how to determine the strength of the cabane/fuselage attachments particularly with regard to the drag truss. I've calculated the vertical reactions at the wing root and found them to be relatively small, most of the lift is reacted by the lift struts. All strut/spar connections are pinned. The pics show the pertinent details. I'll post more if needed.

frontqtr2.jpgfuseside.jpgcabanechannelbracket.jpgwingrig3.jpgaileroncontrol.jpgwingrig2.jpg
 
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Autodidact

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+1 on Joe Fisher's comment, and I prefer a double shear fitting for the main "lift" cabane even if the force there is negligible. There will be side forces creating a moment there from aileron inputs also:

double shear.jpg
 

BobDaly

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Hey Joe,

My nephew's name is Joe Fisher:). The carry-thru is shown in the pic with the control stick. It's the aluminum u-channel running from the angle brackets under the ply floor board.
 

Autodidact

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The strut carry-through would have been better on the outside, but since you've already done it this way, doubling the angle plates that you have at the ends of the strut carry-through with additional angle plates on the outside of the fuselage would be very helpful. You did ask for critiques.

The workmanship looks nice.
 

BobDaly

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Autodidact,

Thanks for the replies. The carry-thru is per Himax plans and I'm not aware of any problems with the Himax carry-thru as I have it. The undercarriage attachment would need to be changed to incorporate your suggestion and this has, in fact, been done by some Himax builders. Your suggestion for the cabane attachment is a good one also.
 

Autodidact

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The carry-thru is per Himax plans...
Ah, that makes me feel good. It is one thing to speculate about "good" practice, and another to back something up with construction according to a designer approved scheme. Lots of Himax out there working OK. I looked at pics of the Minimax and noted that the wing struts attach to the wing at the centroid of the spanwise lift and so, theoretically, there should be basically no vertical force due to lift on the cabane. Also, at max g-force with maximum angle of attack, the drag component of the lift vector should be slightly negative (or is that positive :D), i.e., pushing forward a little. If your cabane attachments are patterned after the Himax plans, I would think that you should be alright there as well. Nice little plane. Please post pics of it when your finished, I've never seen a "Paramax"!
 
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BobDaly

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Autodidact,

"Paramax", I like it. "the drag component of the lift vector should be slightly negative ", This is the heart of my reason for posting. I know that the drag component for the lift vector at 4g and high alpha is 214lbs acting forward. My questions are, "does my cabane structure have to react this entire force which would resolve to a ~300 lbs compressive force in each angled strut?" and "what about the low alpha, cruise drag condition where, I assume, the drag would be reward acting?"

And thanks for confirming that "the wing struts attach to the wing at the centroid of the spanwise lift and so, theoretically, there should be basically no vertical force due to lift on the cabane" which was the result of my calculations.
 

autoreply

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"Paramax", I like it. "the drag component of the lift vector should be slightly negative ", This is the heart of my reason for posting. I know that the drag component for the lift vector at 4g and high alpha is 214lbs acting forward. My questions are, "does my cabane structure have to react this entire force which would resolve to a ~300 lbs compressive force in each angled strut?" and "what about the low alpha, cruise drag condition where, I assume, the drag would be reward acting?"
Lift doesn't exist actually ;-)

A wing generates a single force. Usually we assume that force at the 25% chord point. If the total force doesn't act through that point we now have a single force, a direction of that force and a moment, because we're using the 25% chord point, instead of where the aerodynamic forces really work.

Let's ignore the moment. Now we have a single force. That one, we split up in two (theoretical) ones, lift and drag.

By definition, lift acts perpendicular to the airflow and drag acts parallel to the airflow. If the chord (line from LE to TE) is aligned with the airflow, we can use those theoretical forces (lift and drag) and solve them for our structural calculations.


Now assume we are flying with 30 degrees angle of attack in a stable, horizontal flight, which gives us 100 lbs of lift and 5 lbs of drag on the wing. By definition, the lift acts vertical and the drag horizontal. But our wing and thus the structure is canted backwards 30 degrees and they are loaded rather different.

If we now convert our aerodynamic forces to our structure and it's canted layout we get:
Fpar= aerodynamic forces, parallel to the chord, rearward is positive
Fper=aerodynamic forces, perpendicular to the chord (plain spar bending actually), upwards is positive

Fpar = Drag*COS(30)+Lift*-SIN(30)
Take note that the lift component works negative (forward)

Fper=Lift*COS(30)+Drag*SIN(30)

If we do the math, Fpar is indeed negative and thus points forward:
Fpar = Drag*COS(30)+Lift*-SIN(30)=5*0.86+100*-0.5=4.3-50=-45.7 lbs!

Thus, we have almost half (!) the total lift, working to bend the wing which we perceive as forward. Normally, this isn't that extreme, but those fwd forces can be very high and certainly need to be taken into account.
 

Autodidact

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does my cabane structure have to react this entire force which would resolve to a ~300 lbs compressive force in each angled strut?
If your calculations are correct, then the v-struts would offer no resistance to the negative drag component; without the cabane, the wing would just rotate about the strut carry-through axis.

what about the low alpha, cruise drag condition where, I assume, the drag would be reward acting?
Here, if the strut can handle 300 lb compression, it will certainly handle that much (at least, probably more) tension. That is the strut itself, though, and not necessarily the attachment fitting. If that fitting is antwhere near as strong in tension as the v-strut carry-though is, then you're probably good.

You seem comfortable with math, if you need a buckling formula just ask, some of the guys here ( like Autoreply) can just reel it of the top of their heads. I'd have to look it up, though.:ermm:

By the way, that 300 lb compression load will cause a tension load (almost as high) in the vertical struts. I bet you knew that, though.
 
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BobDaly

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If your calculations are correct, then the v-struts would offer no resistance to the negative drag component; without the cabane, the wing would just rotate about the strut carry-through axis.
That is what I suspected. My calculated compression in the slanted struts is based upon a description of a static test provided in a stress analysis report produced by the designer (It was more of a narative with results and did not show the actual calculations). The static test involved inverting the aircraft with the nose angled 7° down and loading the wings with 1760lbs of sand bags which represents a 4.4g load. Thus the drag force = 1760(sin7) or 214.5 lbs. Is this correct or am I all wet?


If that fitting is antwhere near as strong in tension as the v-strut carry-though is, then you're probably good.
Good point.

You seem comfortable with math...By the way, that 300 lb compression load will cause a tension load (almost as high) in the vertical struts. I bet you knew that, though.
Well, I have a 30 year old piece of paper (BA, Math) that should testify to that but I get nervous:para: everytime I get correspondence from my university, I imagine they're writing to tell me it was all a big mistake, I was strictly a 'C' student. I'm actually glad to find they just want money:gig:

Also, thanks Autoreply for your explanation. The diagram is especially helpful.
 
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Autodidact

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The static test involved inverting the aircraft with the nose angled 7° down and loading the wings with 1760lbs of sand bags which represents a 4.4g load. Thus the drag force = 1760(sin7) or 214.5 lbs. Is this correct or am I all wet?
The math is correct, but, what was angled at 7°? The airfoil nose or the airplanes nose? There is a couple or three deg difference there. So you could be looking at as much as 1760(sin10). Usually, high speed stall occurs at a Clmax that is close to (and maybe more than) the SECTION Clmax due to "dynamic overshoot", i.e., it takes an instant or two for the air to remember that it is supposed to let go.

1760/4.4 = 400lb, is this your gross wt?

As an example, the last (only, to be honest :whistle:) aircraft that I calculated the chordwise forces on was an ultralight with Wmax of 500 lb and 119 ft² wing area. I fantasized that I could make this strong enough for aerobatics w/o busting the pt 103 weight limit. At 6g, the max chordwise force I calculated was 267 lb for ONE wing panel. The normal force was 1241 lb for one panel, so the "lift" vector would be √(1241² + 267²) = 1269 lb for one panel. 500 * 6 = 3000 lb. This is for the wing panel only and does not include the wing area passing through the fuselage and also includes inertial relief for the weight of the wing panel. So, on my design, the center section or cabane (if it had one, it was actually a mid-wing), would need to withstand 534 lb forward force. This is for a heavier plane at substantially more g loading than yours. These calcs were done for an alpha of 19.75 degrees. Dynamic overshoot and a conservative outlook are responsible for this high angle.

For structural design purposes, it is always good to be conservative, but not overly so. For example, from Hiscock's Design of Light Aircraft:

"Chordwise loads, forwards, result from the chordwise component of the lift being greater than that of the wing drag component. At low angle of attack, any rearward component is too low to provide a design case.
However, drag loads should not be ignored. An arbitrary choice of a compareable rearward load will not produce a significant weight penalty."

Profile drag coefficients can be had from the airfoils drag polar chart, i.e., Cd plotted against Cl. The Minimax airfoil is a modified airfoil that Mr. Ison created and I gather that it has high drag at high alpha. Possibly similar to NACA 4415 or maybe even a little more.

Induced drag can be calculated by by Di = kL²/(pi*q*b²) where k=1.35 to 1.4 (it's 1.3 for a J-3 Cub) L is lift, q is mph²/391 (the standard Bernoulli equation rearranged) and b is the wing span in ft.

Add profile drag and induced drag, multiply that sum by cos(alpha), subtract from lift*sin(alpha) and this should give you your chordwise "forward" force.

Remember, the "static" test did not include drag force, so when all is said and done, your calculations might not be much different from the results of the static test, but I would encourage you to do these calcs anyway.
 
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BobDaly

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Ok. I put all that on and walked around in it as my old physics professor used to say and came up with some figures. But to answer your questions, I believe the static test was only to the 'normal category' load limit. The wing panels weigh about 40lbs each so the stress on the airframe was 1760+80 or 1840 and 1840/3.8 ~ 480lbs. And the nose of the aircraft was angled down 7° but the wing incidence is 2° so I should use 9° like you suggested.

Now, I calculate 1840sin9 = 288 lbs chordwise, forward.

Then,

Calculating profile drag I have: .5*.00237*132²*108*.05 = 112 lbs ( I used 132ft/s which is Vne, 90mph, and a drag coefficient of .05 for the airfoil which is a slightly stretched NACA4415)

And induced drag: 1.4(1840²)/3.14*(90²/391)*24² = 126 lbs

So total drag = 112 + 126 = 238 lbs

and

chordwise force = 1840sin16 - 238cos16 = 278 lbs ( Is 16° realistic? The stall angle for NACA4415 is 14°)
 
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Autodidact

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...put all that on and walked around in it as my old physics professor used to say...
I like that. A reminder that we're somewhat at the mercy of our tailor!

That this came out so close to the static test results suggests that this might just be an exercise in confidence building. It looks like you're about to zero in on it. The clincher would be to get good data for the Ison airfoil and the way to do that would be to get an accurate plot of it and the NACA foil and compare them with Xfoil or some other good program, note the differences in Cd and CLmax and apply those differences to the actual wind tunnel results of the NACA airfoil. I suspect that the Ison airfoil has more drag and a slightly lower CLmax than the NACA 44 series; if true, then that would tend to reduce the forward force (more drag, smaller alpha).

But if the static test results ARE what Ison used to proof load his design; I would be tempted to go with that ( the 288 lb), calculate the compression load in the struts and remember to use a factor of safety of at least 1.5. The force needed to buckle that tube is less than that needed to pull it apart, so if it will stand the compression load, then your probably done.

Please bear in mind that I am NOT an engineer and can only regurgitate what I THOUGHT it was that I read :whistle:.
 

Autodidact

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Bob, normally, max force occurs at point "A" on the envelope. Point "A" is the speed at which stall and limit g coincide, i.e., maneuvering speed. First you have to find your modified CLmax for 3d wing effects; 2d Cl/alpha is approximately 0.1/deg for all normal airfoils up to 18% thick or so. Modified Cl/alpha is a = 0.1*R/(R+2) where R is the aspect ratio. TOWS has CLmax for 4415 section occuring at alpha = 15°. So 3d CLmax is 15°*a; then add about 0.2 (someone correct me?) for dynamic over shoot and you should be back up to around 1.3 or so. Then, mph = √((L*391)/(area*CLmax)).

From there you can calculate wing drag and lift as before. I got a chordwise force of 309 lb; still not far from the 288 lb.

Oh, and since the dynamic CLmax is 1.3, the dynamic alpha is 1.3/a.
 
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BobDaly

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I would be tempted to go with that ( the 288 lb), calculate the compression load in the struts and remember to use a factor of safety of at least 1.5. The force needed to buckle that tube is less than that needed to pull it apart, so if it will stand the compression load, then your probably done.

Please bear in mind that I am NOT an engineer and can only regurgitate what I THOUGHT it was that I read :whistle:.
Ok, gotcha. According to Euler, the buckling strength of my 1" x 0.058" wall x 25" long AL tubing is 3000 lbs so no problem there. I'm contemplating gluing a diagonal wood member running from the ply doubler under the forward strut angle bracket down to the verticle member at the seat support like so:
compressmbr.jpg

In order to brace against the compression load.
 

Autodidact

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I'm contemplating gluing a diagonal wood member running from the ply doubler under the forward strut angle bracket down to the verticle member at the seat support like so:
That's exactly what I was going to suggest, since you exhaust pipe support prevents going to the lower longeron.

By the way, I calculated that a 3ft, 1" diameter, 1/64" wall 6061-t6 tube would support about 700 lb or so before buckling, again according to Euler, and you're far in excess of that.

Please let us know what differences you find in handling and performance - I would especially like to know the takeoff run on grass!
 
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