Original Posted By: Mike Tunnicliffe
Pietenpol-List: Wing drag cables and fittings
Pietenpol-List: Wing drag cables and fittings
Original Posted By: Michael Perez
RE: Pietenpol-List: Wing drag fitting math
Original Posted By: owner-pietenpol-list-server(at)matronics.com [mailto:owner-pietenpol-lis
If I did the math right with the right formulas, would aluminum be a good substitute for these fittings? The other drag wire fittings are 3/4" wide. If all were made 3/4" wide, then the rating is closer to 3,000 lbs. -I would be concerned with hole elongation of the fitting. Anyone know how to figure out THOSE numbers?- Thanks all. -It depends on the edge distance from the hole to the edge of the piece.- Fittings like this typically fail in shear ' the bolt will plull sideways through the metal fitting, shearing out a plug of material the width of the bolt and the length of whatever edge distance the fitting was made with.- Typically, the edge distance should be twice the hole diameter (edge distance being measured from the edge of the part to the centerline of the hole) or more.- In this case, using your example of .125=94 aluminum, with a #10 bolt (AN3) the edge distance would be 3/8=94.- After subtracting half the bolt diameter (since that distance was measured to the centerline of the bolt), you are left with .375=94 - .188=94/2, or .281=94.- This gives a shear area of .281=94 x .125=94 (material thickness) or .035 sq. inches of shear area.- Since most metals will fail in shear at about half their tensile strength, the actual failure will occur when the metal reaches a shear stress of 22,500 psi (half of the 45,000 psi you have in tensile strength).- This means that the load required for the fitting to fail in shear is 22,500 x .035 sq in., or 790 lbs.- Note that this also does not apply ANY safety factor.- Of course, any scratch or imperfection will induce a stress concentration making the actual failure load even lower than this.- This is why you should always have a safety factor of at least 1.5 applied to any engineering loading-It is generally good design to make the fittings and hardware at least as strong as the cable so that if the plane is overloaded, the cable will stretch a bit, rather than have parts fail that are more difficult to replace.-And don=92t think that the only loads those drag wire and anti-drag wires will see are the initial installation loads.- The primary purpose of the drag wires is to resist the wind loads on the wing.- The force of the wind produced by airspeed is generally calculated by the formula-- F=AxPxCd, where A is the frontal area in sq. feet, P is the wind pressure in psf and Cd is the drag coefficient.- For the sake of argument, accept that generally wind pressure is taken as 0.00256 V2, where V is the airspeed in mph.- Cd is 2.0 for a flat plate, for a high drag airfoil like a Pietenpol let=92s assume that at worst case (High angle of attack) it is about .5 (makes the calculations easier).- If you just take the frontal area of the wing it is roughly 6=94 thick and 13=92 (per panel) long for a total area of 6.5 sq ft.- At 100 mph, with a Cd of 0.5, this would give a total drag force per wing panel of F = 6.5 x .00256 (100)x (100) x .5 or 83.2 lbs.- Not much load.- However, while Bernard Pietenpol was a genius on most things, he didn=92t do us any favors with the design of the drag wires.- Most planes space the fittings for drag wires about as far apart as the spars are separated, so the angle of the drag wires to the spars is about 45 degrees.- On the Pietenpol, there are only two bays per panel, so the angle is pretty sharp ' about 19 degrees.- This means that for the cables to resist an 83 lb load chordwise load, the load in the cable is actually 254 lbs (F = Load/sine 19deg).- Note that this 254 lb load is in addition to whatever tensile load is present in the cable at rest.- This is why Gene Rambo=92s comment to not make the cable any tighter than necessary is good advice.- It doesn=92t take much preloading to get the cable up to several hundred pounds of tension, and then the flight loads can exceed the strength of the drag wires, and the wing will fold up.-Note also that increasing the speed from 100 mph to 110 causes the drag load to go from 83 lbs to 100 lbs per panel, which causes the cable load to go from 254 to 309 lbs.- And this is assuming that all the drag of the wing is produced by the airfoil, but all the drag of the lift struts and the bracing wires is also carried by the drag wires(and to a small extent by the flying wires between the struts), so the actual loads are higher than shwon.-Sorry ' long answer to why you don=92t want to use aluminum for your wing fittings.- It also should convince you that you really want 1/8=94 cable here and in any high load areas, and use 3/32=94 for areas of lesser load.- 3/32=94 cable has less than half the strength of 1/8=94.-Jack PhillipsNX899JP-
If I did the math right with the right formulas, would aluminum be a good substitute for these fittings? The other drag wire fittings are 3/4" wide. If all were made 3/4" wide, then the rating is closer to 3,000 lbs. -I would be concerned with hole elongation of the fitting. Anyone know how to figure out THOSE numbers?- Thanks all. -It depends on the edge distance from the hole to the edge of the piece.- Fittings like this typically fail in shear ' the bolt will plull sideways through the metal fitting, shearing out a plug of material the width of the bolt and the length of whatever edge distance the fitting was made with.- Typically, the edge distance should be twice the hole diameter (edge distance being measured from the edge of the part to the centerline of the hole) or more.- In this case, using your example of .125=94 aluminum, with a #10 bolt (AN3) the edge distance would be 3/8=94.- After subtracting half the bolt diameter (since that distance was measured to the centerline of the bolt), you are left with .375=94 - .188=94/2, or .281=94.- This gives a shear area of .281=94 x .125=94 (material thickness) or .035 sq. inches of shear area.- Since most metals will fail in shear at about half their tensile strength, the actual failure will occur when the metal reaches a shear stress of 22,500 psi (half of the 45,000 psi you have in tensile strength).- This means that the load required for the fitting to fail in shear is 22,500 x .035 sq in., or 790 lbs.- Note that this also does not apply ANY safety factor.- Of course, any scratch or imperfection will induce a stress concentration making the actual failure load even lower than this.- This is why you should always have a safety factor of at least 1.5 applied to any engineering loading-It is generally good design to make the fittings and hardware at least as strong as the cable so that if the plane is overloaded, the cable will stretch a bit, rather than have parts fail that are more difficult to replace.-And don=92t think that the only loads those drag wire and anti-drag wires will see are the initial installation loads.- The primary purpose of the drag wires is to resist the wind loads on the wing.- The force of the wind produced by airspeed is generally calculated by the formula-- F=AxPxCd, where A is the frontal area in sq. feet, P is the wind pressure in psf and Cd is the drag coefficient.- For the sake of argument, accept that generally wind pressure is taken as 0.00256 V2, where V is the airspeed in mph.- Cd is 2.0 for a flat plate, for a high drag airfoil like a Pietenpol let=92s assume that at worst case (High angle of attack) it is about .5 (makes the calculations easier).- If you just take the frontal area of the wing it is roughly 6=94 thick and 13=92 (per panel) long for a total area of 6.5 sq ft.- At 100 mph, with a Cd of 0.5, this would give a total drag force per wing panel of F = 6.5 x .00256 (100)x (100) x .5 or 83.2 lbs.- Not much load.- However, while Bernard Pietenpol was a genius on most things, he didn=92t do us any favors with the design of the drag wires.- Most planes space the fittings for drag wires about as far apart as the spars are separated, so the angle of the drag wires to the spars is about 45 degrees.- On the Pietenpol, there are only two bays per panel, so the angle is pretty sharp ' about 19 degrees.- This means that for the cables to resist an 83 lb load chordwise load, the load in the cable is actually 254 lbs (F = Load/sine 19deg).- Note that this 254 lb load is in addition to whatever tensile load is present in the cable at rest.- This is why Gene Rambo=92s comment to not make the cable any tighter than necessary is good advice.- It doesn=92t take much preloading to get the cable up to several hundred pounds of tension, and then the flight loads can exceed the strength of the drag wires, and the wing will fold up.-Note also that increasing the speed from 100 mph to 110 causes the drag load to go from 83 lbs to 100 lbs per panel, which causes the cable load to go from 254 to 309 lbs.- And this is assuming that all the drag of the wing is produced by the airfoil, but all the drag of the lift struts and the bracing wires is also carried by the drag wires(and to a small extent by the flying wires between the struts), so the actual loads are higher than shwon.-Sorry ' long answer to why you don=92t want to use aluminum for your wing fittings.- It also should convince you that you really want 1/8=94 cable here and in any high load areas, and use 3/32=94 for areas of lesser load.- 3/32=94 cable has less than half the strength of 1/8=94.-Jack PhillipsNX899JP-
Re: Pietenpol-List: Wing drag cables and fittings
Original Posted By: Michael Perez
Yes, those calculations are right BUT that's not the place tocalculate from. Remember, you have a bolt holding thefitting and strut together. That's a hole in the strap.If the strap is 1" wide and the hole is 1/4" for instance, thenthe material you have to work with is 3/4". So your strengthis now 0.75 X 0.080 X 97000.Any failure is going to be at the weakest point. One of thefour 3/8" sides at the holes. An example of a weak point is a nick at the edge of a hole. That's why it's important toease the edges by deburring them and rounding them inthe process.Each of those four outer strut connections takes a portionof the entire weight of the airplane. The front ones takemore than the rear. A 1200 lb aircraft in level flight haseach wing panel holding up 600 lb. Depending on the airfoil and location of the spars, the front one is likely tosee 325 to 375 lb. That's upward force at 90=B0 to thewing surface. The strut is at an angle so the tension onit is quite a bit more than that upward force. Now thewing is attached to the rest of the plane by both the outerstruts and, in our case, the inner struts. The inner onestake only a small portion of the load though. How much depends on the exact location of the outer ones, the aspect ratio and the planform of the wing, is it Hershy Bar, Tapered, Elliptical, etc..Once you have figured out that static load then you needto take into account turning loads,gust loads etc.Taking the above figures, the strap, at the hole location, hasa tensile strength of 5820 lb. The two straps then, 11640 lb.Quite a bit stronger than needed in our application whenusing 4130. You must keep in mind the Piet was originalydesigned for, and built with mild steel. In fact a lot of certifiedsmall aircraft had mild steel in their frames.Clif ----- Original Message -----
Yes, those calculations are right BUT that's not the place tocalculate from. Remember, you have a bolt holding thefitting and strut together. That's a hole in the strap.If the strap is 1" wide and the hole is 1/4" for instance, thenthe material you have to work with is 3/4". So your strengthis now 0.75 X 0.080 X 97000.Any failure is going to be at the weakest point. One of thefour 3/8" sides at the holes. An example of a weak point is a nick at the edge of a hole. That's why it's important toease the edges by deburring them and rounding them inthe process.Each of those four outer strut connections takes a portionof the entire weight of the airplane. The front ones takemore than the rear. A 1200 lb aircraft in level flight haseach wing panel holding up 600 lb. Depending on the airfoil and location of the spars, the front one is likely tosee 325 to 375 lb. That's upward force at 90=B0 to thewing surface. The strut is at an angle so the tension onit is quite a bit more than that upward force. Now thewing is attached to the rest of the plane by both the outerstruts and, in our case, the inner struts. The inner onestake only a small portion of the load though. How much depends on the exact location of the outer ones, the aspect ratio and the planform of the wing, is it Hershy Bar, Tapered, Elliptical, etc..Once you have figured out that static load then you needto take into account turning loads,gust loads etc.Taking the above figures, the strap, at the hole location, hasa tensile strength of 5820 lb. The two straps then, 11640 lb.Quite a bit stronger than needed in our application whenusing 4130. You must keep in mind the Piet was originalydesigned for, and built with mild steel. In fact a lot of certifiedsmall aircraft had mild steel in their frames.Clif ----- Original Message -----
Re: Pietenpol-List: Wing drag cables and fittings
Original Posted By: Michael Perez
________________________________________________________________________________Date: Mon, 15 Dec 2008 08:40:53 -0800 (PST)
________________________________________________________________________________Date: Mon, 15 Dec 2008 08:40:53 -0800 (PST)
RE: Pietenpol-List: Wing drag cables and fittings
Original Posted By: "Jack T. Textor"
RE: Pietenpol-List: Wing drag cables and fittings
Original Posted By: "Phillips, Jack"
RE: Pietenpol-List: Wing drag cables and fittings
Original Posted By: Michael Perez
RE: Pietenpol-List: Wing drag cables and fittings
Original Posted By: Michael Perez
RE: Pietenpol-List: Wing drag fitting math
Original Posted By: "Phillips, Jack"
Pietenpol-List: Wing drag fitting math
Original Posted By: Michael Perez
Pietenpol-List: Wing drag fitting math
Original Posted By: Michael Perez
RE: Pietenpol-List: Wing drag cables and fittings
Original Posted By: owner-pietenpol-list-server(at)matronics.com
Gene,Let's just say that the plane is "safely overbuilt". Making a statement like"grossly overbuilt" could lead someone who does not possess the necessaryknowledge to assume that they could safely reduce the size of almost any (orall) component(s) of the plane. This simply is not the case. Since the planswere drawn in an age where mild steel was the norm for aircraftconstruction, and today considerably stronger 4130 is the norm, there aresome areas where there is potential to safely modify certain components(again, assuming that one has the requisite knowledge to do the necessarycalculations). But, in general, if built to the plans, everything shouldwork fine. I agree that the drag/anti-drag wires do not carry much load -when the plane is sitting on the ground. When in flight (particularily atthe upper end of the speed range), the draggy old Piet wing may very wellwitness significant loads. The cables wouldn't be where they are, and sizedas they are for no reason. Your point about not overtightening the wires isa very good one. The cables are there to carry loads imposed on them by thedrag induced by flight. No sense in pre-stressing the wires or the airframe.They should be snug, but not overtightened.As for your statements regarding the load distribution on the lift struts,instead of drawing "wrath", all you're going to draw from me is "math". Yourexample of a steel beam being lifted with a cable attached to each end is aninteresting one. Let's assume the beam weighs 1000 pounds. In an idealworld, the cables attached to each end of the beam would be infinitely long,which would result in the cables running vertically, so there would be zerohorizontal component to the load. Since all that leaves is the verticalload, each cable takes half, or 500 pounds. Unfortunately, most of us livein the real world, and we cannot use infinitely long cables. In order tomake use of limited space, the cables are shortened, resulting in atriangular set-up, as you mentioned. When the arrangement changes fromvertical cables to angled cables, horizontal loads are imposed. The verticalcomponent remains at 1000 pounds (due to gravity), but the horizontalcomponent increases with each degree that the cable moves off of vertical.These horizontal forces are ADDED to the vertical forces, to give aresultant force, which acts along the length of the cable. Typically, inpractice, the angle formed between the beam and the cable is not less than30 degrees. At 30 degrees, the resultant force acting on the cable isexactly twice the vertical component. For this reason, each of the cablesused to lift a 1000 pound beam needs to be rated for 1000 pounds, since eachend will actually be loaded at 1000 pounds. If the angle is less than 30degrees, the load goes up even higher. For instance, if the angle is only 20degrees, the resultant load will be almost 1500 pounds. If the angle is only10 degrees, the resultant load is almost 3000 pounds. These calculations arenot complicated, just using basic trigonometry (sine).So, in practical terms, if you're lifting a beam with cables arranged at 30degrees, each of the cables does need to be strong enough to carry the fullweight of the beam, since that IS the (resultant) load each cable will beloaded to. BUT... each of the cables is still only carrying half the load -the difference is that the resultant load is actually twice the weight ofthe beam (in this case).I've attached a simple sketch showing the three cases mentoned above (30, 20and 10 degrees). By carefully drawing accurate triangles to scale, it ispossible to calculate the resultant loads without dredging up unpleasant(for many) memories of high school trigonometry.Bill C. _____
Gene,Let's just say that the plane is "safely overbuilt". Making a statement like"grossly overbuilt" could lead someone who does not possess the necessaryknowledge to assume that they could safely reduce the size of almost any (orall) component(s) of the plane. This simply is not the case. Since the planswere drawn in an age where mild steel was the norm for aircraftconstruction, and today considerably stronger 4130 is the norm, there aresome areas where there is potential to safely modify certain components(again, assuming that one has the requisite knowledge to do the necessarycalculations). But, in general, if built to the plans, everything shouldwork fine. I agree that the drag/anti-drag wires do not carry much load -when the plane is sitting on the ground. When in flight (particularily atthe upper end of the speed range), the draggy old Piet wing may very wellwitness significant loads. The cables wouldn't be where they are, and sizedas they are for no reason. Your point about not overtightening the wires isa very good one. The cables are there to carry loads imposed on them by thedrag induced by flight. No sense in pre-stressing the wires or the airframe.They should be snug, but not overtightened.As for your statements regarding the load distribution on the lift struts,instead of drawing "wrath", all you're going to draw from me is "math". Yourexample of a steel beam being lifted with a cable attached to each end is aninteresting one. Let's assume the beam weighs 1000 pounds. In an idealworld, the cables attached to each end of the beam would be infinitely long,which would result in the cables running vertically, so there would be zerohorizontal component to the load. Since all that leaves is the verticalload, each cable takes half, or 500 pounds. Unfortunately, most of us livein the real world, and we cannot use infinitely long cables. In order tomake use of limited space, the cables are shortened, resulting in atriangular set-up, as you mentioned. When the arrangement changes fromvertical cables to angled cables, horizontal loads are imposed. The verticalcomponent remains at 1000 pounds (due to gravity), but the horizontalcomponent increases with each degree that the cable moves off of vertical.These horizontal forces are ADDED to the vertical forces, to give aresultant force, which acts along the length of the cable. Typically, inpractice, the angle formed between the beam and the cable is not less than30 degrees. At 30 degrees, the resultant force acting on the cable isexactly twice the vertical component. For this reason, each of the cablesused to lift a 1000 pound beam needs to be rated for 1000 pounds, since eachend will actually be loaded at 1000 pounds. If the angle is less than 30degrees, the load goes up even higher. For instance, if the angle is only 20degrees, the resultant load will be almost 1500 pounds. If the angle is only10 degrees, the resultant load is almost 3000 pounds. These calculations arenot complicated, just using basic trigonometry (sine).So, in practical terms, if you're lifting a beam with cables arranged at 30degrees, each of the cables does need to be strong enough to carry the fullweight of the beam, since that IS the (resultant) load each cable will beloaded to. BUT... each of the cables is still only carrying half the load -the difference is that the resultant load is actually twice the weight ofthe beam (in this case).I've attached a simple sketch showing the three cases mentoned above (30, 20and 10 degrees). By carefully drawing accurate triangles to scale, it ispossible to calculate the resultant loads without dredging up unpleasant(for many) memories of high school trigonometry.Bill C. _____
Re: Pietenpol-List: Wing drag cables and fittings
Original Posted By: "Ed G."
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RE: Pietenpol-List: Wing drag cables and fittings
Original Posted By: "Jack T. Textor"
Re: Pietenpol-List: gas tank filler neck options
Original Posted By: John Egan
John:Sory for the dealyed reply as we have been without power here for a few days.I have the Wag Aero filler neck in aluminum cat # E-448-000 and it is 1.25 inch tall. I also have a mild steel one cat # E-323-000 and is the same height and is threaded and fits in to flange Cat # E-322-000Michael in Maine ----- Original Message -----
John:Sory for the dealyed reply as we have been without power here for a few days.I have the Wag Aero filler neck in aluminum cat # E-448-000 and it is 1.25 inch tall. I also have a mild steel one cat # E-323-000 and is the same height and is threaded and fits in to flange Cat # E-322-000Michael in Maine ----- Original Message -----
Pietenpol-List: gas tank filler neck options
Original Posted By: Michael Perez
pol-list(at)matronics.com> RE: Pietenpol-List: Wing drag cables and fi
Original Posted By: owner-pietenpol-list-server(at)matronics.com [mailto:owner-pietenpol-lis
Hi John I'm not sure what other builders have done with their compression struts but I left mine free until after I did a pre-trammel of the wing panel ( 3 piece wing) and then glued them in with T-88. The airleron has a seperate spar set which is shown in cross section in the lower left hand corner of drawing five of the plans. I hope I understood your question correctly and this is of some help. Ed G.00Sorry guys obviously I am not in a position to offer advice as I am still looking at drawings while I wait for my spar material to come in. Please bare with me for what may appear foolish or premature questions: Are compression struts supposed to be free floating and not securely fastened glued or nailed?The aileron cut: does that cut include a rip of the spar or is there another piece that gets is added that runs along the spar? Where does that part of the aileron come from? For some reason I am not seeing where that part comes from and I don't see a call out for that part in the drawings. What am I missing?ThanksJohnSent from my Verizon Wireless BlackBerrypol-list(at)matronics.com>Subject: RE: Pietenpol-List: Wing drag cables and fittingsJack=92s tip on holding the compression struts in place reminded me of what I did on mine. I made little plywood sockets out of 1/8=94 plywood=2C just large enough for the end of the compression strut ti fit inside the socket. I glued the sockets in place on the spar where the compression struts would go. The struts then just nest inside the sockets and are held in place by the compression applied by the drag and anti-drag wires. No nails or glue hold the struts in place. The sockets were just a way to hold them until the wires were tensioned=2C and to keep them from slipping sideways due to shock loads. I don=92t have a good picture showing exactly what I did but you can see the end of the bottom right inboard compression strut sitting in its socket in this picture:Jack PhillipsNX899JPRaleigh=2C NC
Hi John I'm not sure what other builders have done with their compression struts but I left mine free until after I did a pre-trammel of the wing panel ( 3 piece wing) and then glued them in with T-88. The airleron has a seperate spar set which is shown in cross section in the lower left hand corner of drawing five of the plans. I hope I understood your question correctly and this is of some help. Ed G.00Sorry guys obviously I am not in a position to offer advice as I am still looking at drawings while I wait for my spar material to come in. Please bare with me for what may appear foolish or premature questions: Are compression struts supposed to be free floating and not securely fastened glued or nailed?The aileron cut: does that cut include a rip of the spar or is there another piece that gets is added that runs along the spar? Where does that part of the aileron come from? For some reason I am not seeing where that part comes from and I don't see a call out for that part in the drawings. What am I missing?ThanksJohnSent from my Verizon Wireless BlackBerrypol-list(at)matronics.com>Subject: RE: Pietenpol-List: Wing drag cables and fittingsJack=92s tip on holding the compression struts in place reminded me of what I did on mine. I made little plywood sockets out of 1/8=94 plywood=2C just large enough for the end of the compression strut ti fit inside the socket. I glued the sockets in place on the spar where the compression struts would go. The struts then just nest inside the sockets and are held in place by the compression applied by the drag and anti-drag wires. No nails or glue hold the struts in place. The sockets were just a way to hold them until the wires were tensioned=2C and to keep them from slipping sideways due to shock loads. I don=92t have a good picture showing exactly what I did but you can see the end of the bottom right inboard compression strut sitting in its socket in this picture:Jack PhillipsNX899JPRaleigh=2C NC