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Author Topic: Engineering problem for a Corsair  (Read 5797 times)

Offline Don Hutchinson AMA5402

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Engineering problem for a Corsair
« on: September 24, 2015, 01:32:06 PM »
One of the profile models in my warbird series is the Vought Corsair. It of course has the gull wings. I know that there will be a considerable upward force at the gull joint during a pull test but not sure how to calculate the actual amount. I have worked the problem from two angles but not sure it is the correct method. I would like a bit of help from some of you mechanical engineers. a solution and method  would be appreciated. The correct solution will get you a gratis set of the PDF files for the model. If it is the same as mine?, amazing!! A sketch of the setup is attached.
I come up with 2 answers, one a Tangent function, the other a sine function. Which is the right one?? Tan = 6.14,  Sin = 5.95, neither??
Sorry about the attachment, needs to be opened with Adobe reader.
« Last Edit: September 27, 2015, 04:23:51 PM by Don Hutchinson AMA5402 »

Online Dave_Trible

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Re: Engineering problem for a Corsair
« Reply #1 on: September 29, 2015, 08:52:35 AM »
Hello Don.  Missed seeing you in Tulsa this time.  My engineering manuals are long gone and I'm too far past trying to come up with a calculation for this.  I do have questions though.  I occasionally tinker with the idea of a full- bodied Corsair with a sidewinder .60- now maybe Retro .76 since I acquired one.  I thought about rollers or a second bellcrank at the gull joint but the potential problems seemed too much.  So my thinking is to strategically mount the bellcrank and flatten the wing enough for a clean shot through with the leadouts and no contact elsewhere.  With that arraignment there would be no upward stress on the gull joint from pull testing or flight pull.  What am I missing?  Do you have lead out contact at the gull joint?

Dave
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Offline Phil Krankowski

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Re: Engineering problem for a Corsair
« Reply #2 on: September 29, 2015, 09:32:59 AM »
As drawn, you need to redraw the problem.  

Split the drawing into two right triangles at your unknown, which is currently the height.  Tension is the new hypotenuses, the force in question is the two unknowns added together.  

The force triangles are exact ratios to the physical dimensions.

I would consider making an arrangement to reduce the deflection as much as possible since this will always rub and the added friction is not really helping anything.

Phil

Offline Don Hutchinson AMA5402

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Re: Engineering problem for a Corsair
« Reply #3 on: September 29, 2015, 10:02:05 AM »
Phil- Could you explain a bit more clearly, with a solution? The hypotenuse is the long side of the triangle so what do I do with it? I have tried to reduce the force as much as I could, I lowered the bellcrank from the center of rib 1 and moved the rollers at the gull joint out 1 rib bay. This made quite a difference in the angles.

Dave- Then it wouldn't be a Corsair anymore! Actually, with a full fuselage and a thick (22 to 25%) airfoil, you could probably snake the control system in with only a small direction change by moving the rollers out one or two rib stations beyond the gull joint. One of your big problems will be how to work the flaps! I did it with two small bellcranks at the gull joints. Makes for an interesting design challenge!

Offline Tim Wescott

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Re: Engineering problem for a Corsair
« Reply #4 on: September 29, 2015, 10:36:26 AM »
Hey Don:

Are you worried about the wing structure holding up, or do you have something else eating you about this?

Is it really going to be a 24 ounce plane?  How big is the thing?

Assuming a magic no-friction slider at the wing joint, I get (15lbf)(tan(16o) + tan(7o)), or 6.14lbf.  I suppose that's a lot of force for a 1.5lb plane -- if you have any junk wings lying around you may want to grab one from a similar-weight plane, support it at tip and root, then pile weight onto the 1/4 span (i.e., the location of the bend) until it breaks.  That should tell you a lot about whether you need to worry about wing strength.

You may have already done this, but to the extent that it's a problem it behooves you to straighten the bend as much as you can within the wing -- in other words, don't place the bend at the middle of the wing ribs; rather, place it as high within the wing as you can without the leadouts getting fouled in the tops of the ribs.  At the same time, place the bellcrank as low in the fuselage as you can without the leadouts getting fouled in the bottoms of the ribs.

I think I'd want to mock this up -- I'd probably use nails and a piece of 1x8 board (surely you have a neighbor with appropriate siding), you can do something classier.  Make sure that for normal in-flight line tensions the friction of the leadouts going around the bend doesn't cause too much stick in the control system.  For that matter, you can use the mock-up to verify the force at the joint -- get the angles right and hold the bend down with a 6-1/4 lb weight.  Putting a bit more than 15 pounds on the leadouts should just start lifting the weight.
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Online Dave_Trible

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Re: Engineering problem for a Corsair
« Reply #5 on: September 29, 2015, 12:54:53 PM »
Yes the flaps.  I figured the secondary  bellcranks too with slightly different  control rates so the flaps would bypass and overlap each other a bit.  That would take a little experimenting to get right I'm sure.  Either that or a stationary section between to allow clearance.  Maybe a double jointed lucky box of some sort.  Oh... makes my head hurt.....

Dave

Maybe a torque tube in the wing with dual output arms..
« Last Edit: September 29, 2015, 02:27:26 PM by Dave_Trible »
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Offline Don Hutchinson AMA5402

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Re: Engineering problem for a Corsair
« Reply #6 on: September 29, 2015, 04:42:16 PM »
Tim- I think we are I am using the right method, Tan 16x15 + Tan 7x15 to find the solution. The model is not 24 ounces, I am using 48 ounces as that is what the Dauntless weighs. I am only solving for one line here. Yes, my concern is having the guts pop up out of the wing surface during the pull test! However, I have sheeted wings out to the rib where the two rollers are and have placed enough structure there to hold up OK. My big concern is someone who doesn't glue well or uses 4# wood for this area! I did mock up the setup back then and was amazed to find out how big a force it was, quite close to what we are calculating. As for placing the leadouts above the middle of the rib, I have gone even further, I moved the rollers out 1 rib past the gull joint so the leadouts are very close to the top of the gull rib. I also set the bellcrank below center at the fuselage.

Dave- that is exactly what I did for the flaps, a small follower bellcrank at each gull joint that drive the two flaps at a slightly different rate so that I can overlap the beveled ends of the flaps and thus never see a gap as they go down and not collide as they go up. I did a mockup when I drew the plans and it seemed to work just fine. The real corsair had three sections of flaps

Offline Tim Wescott

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Re: Engineering problem for a Corsair
« Reply #7 on: September 29, 2015, 05:54:01 PM »
This sounds like it's going to be a cool plane when it's done.
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Offline Don Hutchinson AMA5402

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Re: Engineering problem for a Corsair
« Reply #8 on: September 30, 2015, 09:10:59 AM »
Tim and Dave- Watch your E-mails. I assume the address in your profile here is correct.

Offline Don Hutchinson AMA5402

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Re: Engineering problem for a Corsair
« Reply #9 on: October 01, 2015, 10:56:42 AM »
I now believe the above calculations are not correct. In thinking about the problem I tried three scenarios. Tan of angle x 15, Sin of angle x 15 and a linear angle/90 x15. Using the Tanx15 at 0, force is 15# horizontal, at 45 both the hor and vert forces are each 15# which is not possible and at 90 the vertical force is infinity which will break the lines! thus using the Tan table is not the answer, Using the Sinx15, once again the force at 0 is 15# horizontal, at 45 the the hor and vert forces are 10.6# which is also not possible while at 90, the vertical force is 15#. Using a linear scale at 0 again vert at 0 is 0, at 45 the hor and vert are each 7.5# and at 90 the vert is again 15#, the only solution that works, thus the answer for 16/90 + 7/90x15 is 3.833. 23/90x15 gives the same answer whereas when using trig tables this does not work, therefore I think the linear solution is correct. Plus I like the lower numbers I get!

Offline Tim Wescott

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Re: Engineering problem for a Corsair
« Reply #10 on: October 01, 2015, 03:53:16 PM »
I now believe the above calculations are not correct. In thinking about the problem I tried three scenarios. Tan of angle x 15, Sin of angle x 15 and a linear angle/90 x15. Using the Tanx15 at 0, force is 15# horizontal, at 45 both the hor and vert forces are each 15# which is not possible and at 90 the vertical force is infinity which will break the lines! thus using the Tan table is not the answer, Using the Sinx15, once again the force at 0 is 15# horizontal, at 45 the the hor and vert forces are 10.6# which is also not possible while at 90, the vertical force is 15#. Using a linear scale at 0 again vert at 0 is 0, at 45 the hor and vert are each 7.5# and at 90 the vert is again 15#, the only solution that works, thus the answer for 16/90 + 7/90x15 is 3.833. 23/90x15 gives the same answer whereas when using trig tables this does not work, therefore I think the linear solution is correct. Plus I like the lower numbers I get!

Dayum.  Yes, if you look at your original sketch the 15# is along the length of the hypotenuse, and the push upward on the wing is along the side opposite of the angle -- meaning, use the sine.  I'm sorry I didn't double-check my numbers, but your verification method was spot on.  It doesn't matter too much in this case -- this is the sort of thing that I'd design with a healthy margin, 5.94# isn't that far away from 6.14, and at any rate you would have been designing with the safer number.
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Offline Don Hutchinson AMA5402

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Re: Engineering problem for a Corsair
« Reply #11 on: October 02, 2015, 03:07:41 PM »
I finally did what any good test engineer would do! Run a test. I laid out the pattern on a board, then stood it upright and ran a cord from the "bellcrank" to the leadout guide "pulley in this case" and loaded it with 100 ounces. I then added weight at the 6.8" spot until the cord matched the diagram. It took 41.5 ounces!! .Convert that to 41.5% of 15# and the answer is 6.225#!! Kinda where I was when I started this whole thing! Back when I did the drawings in "07, I put a warning in the text doc to make these parts strong as there was an upward force of 6.38# on each leadout so I must have figured it about right then.

Offline Trostle

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Re: Engineering problem for a Corsair
« Reply #12 on: October 03, 2015, 01:59:54 PM »
Don,

There was a time I think I could have worked out the solution for you.  My force diagram shows the cable anchored at the root, moving past the gull wing angle (over a bar where it is assumed there is no friction) and on to the tip.  The tip will align with the root in line with the direction of the pull, regardless of the amount of deflection represented by the gull wing angle.  The pull along that cable is constant at 15 lb.  The force trying to straighten that dihedral break is considerably less than the tension on the cable and will be in the neighborhood of the calculations you have.

Now to some side comments.  I have had some experience with two different scale swept wing designs where I put the bellcrank in the root of the wing.  In both cases, the leadout position, as dictated by the CG had to be well behind the bellcrank position, the the leadouts coming from the wing at nearly mid span position.  In one case, I used small pulleys to change the direction of the leadouts from down the wing span to the leadout position.  This worked to some degree, but the leadouts still showed excessive wear at the pulleys and one failed at the pulley on a pull test.  The other case, I used short smooth steel rods for the leadouts to move over for the direction change from moving down the span to exit at the leading edge at mid span of the wing.  This worked for scale where not that many flights are planned.  But for stunt, I think the drag over those rods, though not offering much resistance, there was still drag and would not be a good thing for stunt.  If the angle change is not much, then you could live with it, but the amount needed to get past that gull wing might be more than you would want.  Also, the wear of those leadouts would be a concern.  I used the plastic tubing used for flexible pushrod guides to "sheath" the leadouts.  I think this reduced the drag somewhat and probably should extend the wear life of the leadouts, but for a stunt model planned to be used for dozens or hundreds of flights, I would still be concerned.

Now, let's talk about Kirk Mullinex and his Coarsair.  This is the only picture I have of it and does not do it justice.  This is powered by a piped PA 65.  It is a large model somewhere approaching 700 sq in.  (It might be slightly smaller, but not by much.)  Molded shell fuselage, and like all of Kirk's stunters, is light, has a great finish, and flies really well.  I think he mounted his bellcrank at the bottom of the gull wing, with the pushrod to a transfer crank in the center of the wing/fuselage. Kirk is a master machinist and he made his own universal joints to operate the flaps across that gull wing.  And there is no play between those universal joints.  You have to feel the flaps move with no flexing between the joints to believe this was possible.  Thought you might be interested.

Keith

Offline Don Hutchinson AMA5402

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Re: Engineering problem for a Corsair
« Reply #13 on: October 03, 2015, 02:53:55 PM »
I have seen Kirk's Corsair at a Golden State meet some years back. A fine piece of work! As for the Corsair, when I did the drawings, I set the system up to give me the lowest load at the gull joint by moving the bellcrank lower in the center rib and moving the rollers at the gull joint out one rib bay, not having a clue that the load would be more than a few pounds. I later thought I should calculate the load and after figuring out how to do it was amazed to find it 6+ pounds per line. If I were to do it again today, I would mess with the angles and bellcrank position to arrive at a better solution. In the text document that goes with the drawings, I notify the builder to make this area "bullet proof! If anyone ever builds one then maybe we will find out if it works! Even Ford put a lemon once!

Offline Don Hutchinson AMA5402

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Re: Engineering problem for a Corsair
« Reply #14 on: October 06, 2015, 10:24:51 AM »
This is my final answer! Worked out on paper and trig tables. Total force is 6.0396#, biased 9 degrees toward the wingtip.


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