Original Posted By: Jim Boyer
Pietenpol-List: Please Check My Math (alum.wing strut)
RE: Pietenpol-List: Please Check My Math (alum.wing strut)
Original Posted By: owner-pietenpol-list-server(at)matronics.com
Michael, how are you arriving at 1200 lbs per strut at 4 G=92s? I could see that if your struts were vertical (in line with the lift load).They are not, and since they are at an angle from the horizontal, each strutwill see a load equal to the lift load at that point divided by the sine ofthe angle. For most Pietenpols, the angle is about 30=B0 and the sine of 30=B0is .50, so if all four struts are carrying a quarter of the total load(they=92re not ' the front struts tend to carry more than the rears, due tothe position of the center of pressure, and the pressure distributioncurve), and the total load is 4800 lbs, each strut is carrying 1200/.5 or2400 lbs. Without knowing the pressure distribution on the wing, I would assume forsafety=92s sake that the front struts are carrying 2/3 of the load and therear struts only 1/3. If that is the case, then the load at each frontstrut fitting on the wing at 4 g=92s with a 1200 lb gross weight would be 2/3(2400) or 1600 lbs. This is assuming that each wing panel is generatinghalf the lift or 2400 lbs, and the centersection is adding nothing, so againthis is a conservative assumption. If the lift load at the front strut is1600 lbs, the load in the strut itself will be twice that (if the angle is30=B0) or 3200 lbs. Quite a bit different than the 1200 lbs you werecalculating. Jack PhillipsNX899JP =93Icarus Plummet=94Raleigh, NC _____
Michael, how are you arriving at 1200 lbs per strut at 4 G=92s? I could see that if your struts were vertical (in line with the lift load).They are not, and since they are at an angle from the horizontal, each strutwill see a load equal to the lift load at that point divided by the sine ofthe angle. For most Pietenpols, the angle is about 30=B0 and the sine of 30=B0is .50, so if all four struts are carrying a quarter of the total load(they=92re not ' the front struts tend to carry more than the rears, due tothe position of the center of pressure, and the pressure distributioncurve), and the total load is 4800 lbs, each strut is carrying 1200/.5 or2400 lbs. Without knowing the pressure distribution on the wing, I would assume forsafety=92s sake that the front struts are carrying 2/3 of the load and therear struts only 1/3. If that is the case, then the load at each frontstrut fitting on the wing at 4 g=92s with a 1200 lb gross weight would be 2/3(2400) or 1600 lbs. This is assuming that each wing panel is generatinghalf the lift or 2400 lbs, and the centersection is adding nothing, so againthis is a conservative assumption. If the lift load at the front strut is1600 lbs, the load in the strut itself will be twice that (if the angle is30=B0) or 3200 lbs. Quite a bit different than the 1200 lbs you werecalculating. Jack PhillipsNX899JP =93Icarus Plummet=94Raleigh, NC _____
RE: Pietenpol-List: Please Check My Math (alum.wing strut)
Original Posted By: Michael Perez
RE: Pietenpol-List: Please Check My Math (alum.wing strut)
Original Posted By: owner-pietenpol-list-server(at)matronics.com
Some load can certainly be carried by the wires, but that's not what theyare there for, and I sure wouldn't count on them to hold your wing on.Cables have a goodly amount of stretch to them and they won't begin to carrymuch load until they've been stretched enough that the other fittings on thestruts have failed.The purpose of those wires is to keep the wing from racking fore and aft(about the yaw axis) - not to carry the lift loads.Jack PhillipsNX899JP "Icarus Plummet"Raleigh, NC-----Original Message-----
Some load can certainly be carried by the wires, but that's not what theyare there for, and I sure wouldn't count on them to hold your wing on.Cables have a goodly amount of stretch to them and they won't begin to carrymuch load until they've been stretched enough that the other fittings on thestruts have failed.The purpose of those wires is to keep the wing from racking fore and aft(about the yaw axis) - not to carry the lift loads.Jack PhillipsNX899JP "Icarus Plummet"Raleigh, NC-----Original Message-----
Re: Pietenpol-List: Please Check My Math (alum.wing strut)
Original Posted By: "Jim Markle"
My plans call for 3/16" diagonal/xbracing wires.----- Original Message -----
My plans call for 3/16" diagonal/xbracing wires.----- Original Message -----
Re: Pietenpol-List: Re: Tail wheel
Original Posted By: Rick Holland
Subject: Re: Pietenpol-List: Re: Tail wheel
Subject: Re: Pietenpol-List: Re: Tail wheel
Re: Pietenpol-List: Re: Tail wheel
Original Posted By: "TOM STINEMETZE"
Re: Pietenpol-List: Re: Tail wheel
Original Posted By: Rick Holland
Rick, what model Matco tailwheel did you use? I am thinking about a Matco WHLT-6. The specs say it has an operating load of 450#. ----- Original Message -----
Rick, what model Matco tailwheel did you use? I am thinking about a Matco WHLT-6. The specs say it has an operating load of 450#. ----- Original Message -----
Re: Pietenpol-List: Re: Tail wheel
Original Posted By: Rick Holland
I don't have my plans at hand, but I don't recall ANY 3/16" cable being calledfor anywhere in the plans.Bill C.Read this topic online here:http://forums.matronics.com/viewtopic.p ... ______Date: Mon, 10 Jan 2011 16:40:51 -0700Subject: Re: Pietenpol-List: Re: Tail wheel
I don't have my plans at hand, but I don't recall ANY 3/16" cable being calledfor anywhere in the plans.Bill C.Read this topic online here:http://forums.matronics.com/viewtopic.p ... ______Date: Mon, 10 Jan 2011 16:40:51 -0700Subject: Re: Pietenpol-List: Re: Tail wheel
Re: Pietenpol-List: Re: Please Check My Math (alum.wing strut)
Original Posted By: Rick Holland
Subject: Re: Pietenpol-List: Re: Please Check My Math (alum.wing strut)
Subject: Re: Pietenpol-List: Re: Please Check My Math (alum.wing strut)
Re: Pietenpol-List: Re: Please Check My Math (alum.wing strut)
Original Posted By: Bill Church
I also used 3/16 cable for the cabanes, like the plans call for. Also it should be used on the controls, aft of the belcrank.Dan HelsperPoplar Grove, IL.-----Original Message-----
I also used 3/16 cable for the cabanes, like the plans call for. Also it should be used on the controls, aft of the belcrank.Dan HelsperPoplar Grove, IL.-----Original Message-----
Re: Pietenpol-List: Re: Please Check My Math (alum.wing strut)
Original Posted By: helspersew
Woops! I meant 3/32"-----Original Message-----
Woops! I meant 3/32"-----Original Message-----
Re: Pietenpol-List: Please Check My Math (alum.wing strut)
Original Posted By: Michael Perez
=EF=BB=0A=0A =0A#yiv805611046 P {=0AMARGIN:0px;}=0A=0A=0AAttached is the lift distribution chart for hershey bar =0Awings in the aspect ratio=0Arange of the Piet wing. From this you can deduce that =0Athere is only a small=0Aportion of the lift supported by the cabanes. Therefore, =0Athe majority of=0Alift is handled by the struts. It's all in the archives, =0Aback there somewhere.
=0A=C2-=0AAlso, my understanding is that one designs and builds to a =0Asafety factor=0Aof 1.5. So if an AC is placarded at 4G then it will go to =0A6 before failure.=0A=C2-=0AClif=0A=0A =0A It has to do =0A with the distribution of lift.=C2- More lift is generated from the inboard =0A section of the wing than the outboard sections.=C2- As a result there is =0A less torque upward on the outer wing panels than if the lift was equal all of =0A the way across the wing.=C2- This means that there is less force imposed on =0A the lift strut.=C2- Malcolm Morrisonhttp://home.comcast.net/~mmorrison123/=0A What in your =0A math is different then Jack's 3200? Your numbers are quite =0A different?Michael Perez________________________________________________________________________________Date: Tue, 11 Jan 2011 05:39:12 -0800 (PST)
=EF=BB=0A=0A =0A#yiv805611046 P {=0AMARGIN:0px;}=0A=0A=0AAttached is the lift distribution chart for hershey bar =0Awings in the aspect ratio=0Arange of the Piet wing. From this you can deduce that =0Athere is only a small=0Aportion of the lift supported by the cabanes. Therefore, =0Athe majority of=0Alift is handled by the struts. It's all in the archives, =0Aback there somewhere.
=0A=C2-=0AAlso, my understanding is that one designs and builds to a =0Asafety factor=0Aof 1.5. So if an AC is placarded at 4G then it will go to =0A6 before failure.=0A=C2-=0AClif=0A=0A =0A It has to do =0A with the distribution of lift.=C2- More lift is generated from the inboard =0A section of the wing than the outboard sections.=C2- As a result there is =0A less torque upward on the outer wing panels than if the lift was equal all of =0A the way across the wing.=C2- This means that there is less force imposed on =0A the lift strut.=C2- Malcolm Morrisonhttp://home.comcast.net/~mmorrison123/=0A What in your =0A math is different then Jack's 3200? Your numbers are quite =0A different?Michael Perez________________________________________________________________________________Date: Tue, 11 Jan 2011 05:39:12 -0800 (PST)Re: Pietenpol-List: Hole Depth for Piper fork
Original Posted By: Rick Holland
=EF=BB Attached is the lift distribution chart for hershey bar wings in the aspect ratio range of the Piet wing. From this you can deduce that there is only a small portion of the lift supported by the cabanes. Therefore, the majority of lift is handled by the struts. It's all in the archives, back there somewhere. Also, my understanding is that one designs and builds to a safety factor of 1.5. So if an AC is placarded at 4G then it will go to 6 before failure. Clif It has to do with the distribution of lift. More lift is generated from the inboard section of the wing than the outboard sections. As a result there is less torque upward on the outer wing panels than if the lift was equal all of the way across the wing. This means that there is less force imposed on the lift strut. Malcolm Morrison http://home.comcast.net/~mmorrison123/ What in your math is different then Jack's 3200? Your numbers are quite different? Michael Perez ________________________________________________________________________________Date: Tue, 11 Jan 2011 11:31:32 -0700Subject: Re: Pietenpol-List: Hole Depth for Piper fork
=EF=BB Attached is the lift distribution chart for hershey bar wings in the aspect ratio range of the Piet wing. From this you can deduce that there is only a small portion of the lift supported by the cabanes. Therefore, the majority of lift is handled by the struts. It's all in the archives, back there somewhere.
Also, my understanding is that one designs and builds to a safety factor of 1.5. So if an AC is placarded at 4G then it will go to 6 before failure. Clif It has to do with the distribution of lift. More lift is generated from the inboard section of the wing than the outboard sections. As a result there is less torque upward on the outer wing panels than if the lift was equal all of the way across the wing. This means that there is less force imposed on the lift strut. Malcolm Morrison http://home.comcast.net/~mmorrison123/ What in your math is different then Jack's 3200? Your numbers are quite different? Michael Perez ________________________________________________________________________________Date: Tue, 11 Jan 2011 11:31:32 -0700Subject: Re: Pietenpol-List: Hole Depth for Piper forkRe: Pietenpol-List: Please Check My Math (alum.wing strut)
Original Posted By: "Jack Phillips"
Attached is the lift distribution chart for hershey bar wings in the aspect ratiorange of the Piet wing. From this you can deduce that there is only a smallportion of the lift supported by the cabanes. Therefore, the majority oflift is handled by the struts. It's all in the archives, back there somewhere. :-)Also, my understanding is that one designs and builds to a safety factorof 1.5. So if an AC is placarded at 4G then it will go to 6 before failure.Clif It has to do with the distribution of lift. More lift is generated from the inboard section of the wing than the outboard sections. As a result there is less torque upward on the outer wing panels than if the lift was equal all of the way across the wing. This means that there is less force imposed on the lift strut. Malcolm Morrison http://home.comcast.net/~mmorrison123/ What in your math is different then Jack's 3200? Your numbers are quite different? Michael Perez________________________________________________________________________________
Attached is the lift distribution chart for hershey bar wings in the aspect ratiorange of the Piet wing. From this you can deduce that there is only a smallportion of the lift supported by the cabanes. Therefore, the majority oflift is handled by the struts. It's all in the archives, back there somewhere. :-)Also, my understanding is that one designs and builds to a safety factorof 1.5. So if an AC is placarded at 4G then it will go to 6 before failure.Clif It has to do with the distribution of lift. More lift is generated from the inboard section of the wing than the outboard sections. As a result there is less torque upward on the outer wing panels than if the lift was equal all of the way across the wing. This means that there is less force imposed on the lift strut. Malcolm Morrison http://home.comcast.net/~mmorrison123/ What in your math is different then Jack's 3200? Your numbers are quite different? Michael Perez________________________________________________________________________________
RE: Pietenpol-List: Please Check My Math (alum.wing strut)
Original Posted By: owner-pietenpol-list-server(at)matronics.com
This is spanwise lift. What I was trying to approximate is the chordwisepressure distribution ' that will determine how much load is carried by thefront strut compared to the rear strut. Since the center of pressure islocated very close to the center of gravity (and hopefully a little bitforward of it, unless the plane is too tail-heavy), it is loacted very closeto the front spar, which means the front spar carries the lion=92s share ofthe lift load. My guess was that it carrries 2/3 of the load and the rearspars only 1/3, but this is only a GUESS. It may be carrying much more ofthe load than that. It might be carrying =BE of the load.Because you don=92t know, you look at what has worked in the past. If youwere a real aircraft factory, the prudent thing to do would be to build theplane with turnbuckle fittings, and then intentionally load it until itbreaks and see what that value was. Then divide that value by 1.5 and callthat your max safe G load.In Michael=92s case, the best thing to do is to build his plane with the cheapfittings, then load it to gross weight and subject it to maneuvers whichinduce 6 G=92s and see if he dies in the test. If he doesn=92t die, then it wasOK.Jack PhillipsNX899JP =93Icarus Plummet=94Raleigh, NC _____
This is spanwise lift. What I was trying to approximate is the chordwisepressure distribution ' that will determine how much load is carried by thefront strut compared to the rear strut. Since the center of pressure islocated very close to the center of gravity (and hopefully a little bitforward of it, unless the plane is too tail-heavy), it is loacted very closeto the front spar, which means the front spar carries the lion=92s share ofthe lift load. My guess was that it carrries 2/3 of the load and the rearspars only 1/3, but this is only a GUESS. It may be carrying much more ofthe load than that. It might be carrying =BE of the load.Because you don=92t know, you look at what has worked in the past. If youwere a real aircraft factory, the prudent thing to do would be to build theplane with turnbuckle fittings, and then intentionally load it until itbreaks and see what that value was. Then divide that value by 1.5 and callthat your max safe G load.In Michael=92s case, the best thing to do is to build his plane with the cheapfittings, then load it to gross weight and subject it to maneuvers whichinduce 6 G=92s and see if he dies in the test. If he doesn=92t die, then it wasOK.Jack PhillipsNX899JP =93Icarus Plummet=94Raleigh, NC _____
RE: Pietenpol-List: Please Check My Math (alum.wing strut)
Original Posted By: pietflyr(at)bellsouth.net
The center of pressure moves quite a bit with angle of attack also. At a 0 or slightly negative angle of attack the load may approach 50/50 and at a high angle of attack may approach 95% being carried by the front spar. Center of pressure movement varies with airfoil shape too. Too many variables for me to fool with.Doug DeverIn beautiful Stow Ohio
The center of pressure moves quite a bit with angle of attack also. At a 0 or slightly negative angle of attack the load may approach 50/50 and at a high angle of attack may approach 95% being carried by the front spar. Center of pressure movement varies with airfoil shape too. Too many variables for me to fool with.Doug DeverIn beautiful Stow Ohio
Re: Pietenpol-List: Please Check My Math (alum.wing strut)
Original Posted By: Clif Dawson
As a non-engineer builder, if the lift struts require a rod end that will withstand 8000 lbs, or even 4000, I have a hard time imagining that those little attach plates made out of 12 g steel and held to the fuselage longerons by 3 or 4 3/16th in bolts would withstand that weight. My common sense make me nervous, but it seems to work?Just a comment.Thanks and fly SAFELY,Ray KrauseWaiex 51YX, Jabiru 3300 (1197), Sensenich wood prop, AeroCarb (#2 needle modified), Dynon D-180, Garmin SL 30 NavCom, Garmin 327 transponder, Garmin Aera 560, nav and strobe lights: 236 hrs., building a Sky Scout.... slowly. ----- Original Message -----
As a non-engineer builder, if the lift struts require a rod end that will withstand 8000 lbs, or even 4000, I have a hard time imagining that those little attach plates made out of 12 g steel and held to the fuselage longerons by 3 or 4 3/16th in bolts would withstand that weight. My common sense make me nervous, but it seems to work?Just a comment.Thanks and fly SAFELY,Ray KrauseWaiex 51YX, Jabiru 3300 (1197), Sensenich wood prop, AeroCarb (#2 needle modified), Dynon D-180, Garmin SL 30 NavCom, Garmin 327 transponder, Garmin Aera 560, nav and strobe lights: 236 hrs., building a Sky Scout.... slowly. ----- Original Message -----
Re: Pietenpol-List: Please Check My Math (alum.wing strut)
Original Posted By: "Jim Markle"
I have drilled so many 3/16" holes I have 3/16" on the brain. I, of course, meant 3/32" cables.----- Original Message -----
I have drilled so many 3/16" holes I have 3/16" on the brain. I, of course, meant 3/32" cables.----- Original Message -----
>>> Pietenpol-List: Please
Original Posted By: KM Heide CPO/FAAOP
Re: Pietenpol-List: Please Check My Math (alum.wing strut)
Original Posted By: gliderx5(at)comcast.net
EF Bruhn's book "Analysis and Design of Aircraft Structures" would disagree with your statement that more lift is generated from the inboard section than the outboard sections. That would be true if the wing is tapered, but I'm looking at a diagram of a rectangular wing with spars of equal depth the entire length and the diagram of airload shows the same load on the tip as on the root. ----- Original Message -----
EF Bruhn's book "Analysis and Design of Aircraft Structures" would disagree with your statement that more lift is generated from the inboard section than the outboard sections. That would be true if the wing is tapered, but I'm looking at a diagram of a rectangular wing with spars of equal depth the entire length and the diagram of airload shows the same load on the tip as on the root. ----- Original Message -----
Re: Pietenpol-List: Please Check My Math (alum.wing strut)
Original Posted By: Michael Perez
Michael, I don't want to see even 4 Gs. ----- Original Message -----
Michael, I don't want to see even 4 Gs. ----- Original Message -----
Re: Pietenpol-List: Re: Tail wheel
Original Posted By: Rick Holland
Subject: Re: Pietenpol-List: Re: Tail wheel
Subject: Re: Pietenpol-List: Re: Tail wheel
RE: Pietenpol-List: Please Check My Math (alum.wing strut)
Original Posted By: cncampbell(at)windstream.net
As I understand aerodynamics and lift (limited) the lift pattern or distribution on a hershey bar wing is eliptical which means there is not as much lift generated at the tips. This is also why washout is not needed=2C although many put some in. This would also imply that there is not as much lift generated at the root either. This would mean that most of the lift would be generated in the middle 80% (paredo principle anyone.) From a design simplification standpoint to figure span loading (when designing spar strength) The lift is considered equal along the span. We atmospheric guys design to yeild strength. The Space shuttle is the only thing I know of designed to ultimate strength. In other words we are all splitting hairs as this is not Rocket science. The darn thing has flown for 80+years. Don't mess with a good thing. It works. Unless=2C of course you are an aeronautical engineer. Then go for it.Doug DeverIn beautiful Stow Ohio
As I understand aerodynamics and lift (limited) the lift pattern or distribution on a hershey bar wing is eliptical which means there is not as much lift generated at the tips. This is also why washout is not needed=2C although many put some in. This would also imply that there is not as much lift generated at the root either. This would mean that most of the lift would be generated in the middle 80% (paredo principle anyone.) From a design simplification standpoint to figure span loading (when designing spar strength) The lift is considered equal along the span. We atmospheric guys design to yeild strength. The Space shuttle is the only thing I know of designed to ultimate strength. In other words we are all splitting hairs as this is not Rocket science. The darn thing has flown for 80+years. Don't mess with a good thing. It works. Unless=2C of course you are an aeronautical engineer. Then go for it.Doug DeverIn beautiful Stow Ohio
Re: Pietenpol-List: Please Check My Math (alum.wing strut)
Original Posted By: Owen Davies
Yes, but I was replying to the message referenced fromMalcolm. And the whole thing started with Michael'squestion regarding the lift strut forces. So both this chartand the info your guessing at, which would have to comefrom research info on the Bernard airfoil, are important.We don't have the airfoil info, only that from similiar oneslike the Jenny's. The rear strut is actually quite far forwardMaking it relatively close to the center of pressure. Withthe allowable CG being between 15" and 20" then wecan assume the CP to fall near the center of that range.The front strut is 7" from the LE making it 10 to 11"ahead of the CP. The rear spar is 36" from the LE makingit 18 to 19" from the CP so the front spar takes between10/19ths to 11/18ths of the load. Two thirds is 66% and11/18ths is 0.62, pretty close to 2/3rds. So, Jack, usingyour guess provides a bit of safety factor. Pretty goodguesstimating. :-)Clif"Education: the path from cocky ignorance to miserable uncertainty."- Mark Twain, writer and humoristMy guess was that it carrries 2/3 of the load and the rear spars only 1/3, but this is only a GUESS. It may be carrying much more of the load than that. It might be carrying =BE of the load. In Michael's case, the best thing to do is to build his plane with the cheap fittings, then load it to gross weight and subject it to maneuvers which induce 6 G's and see if he dies in the test. If he doesn't die, then it was OK. Jack Phillips NX899JP "Icarus Plummet" Raleigh, NC ________________________________________________________________________________Date: Tue, 11 Jan 2011 23:19:57 -0500
Yes, but I was replying to the message referenced fromMalcolm. And the whole thing started with Michael'squestion regarding the lift strut forces. So both this chartand the info your guessing at, which would have to comefrom research info on the Bernard airfoil, are important.We don't have the airfoil info, only that from similiar oneslike the Jenny's. The rear strut is actually quite far forwardMaking it relatively close to the center of pressure. Withthe allowable CG being between 15" and 20" then wecan assume the CP to fall near the center of that range.The front strut is 7" from the LE making it 10 to 11"ahead of the CP. The rear spar is 36" from the LE makingit 18 to 19" from the CP so the front spar takes between10/19ths to 11/18ths of the load. Two thirds is 66% and11/18ths is 0.62, pretty close to 2/3rds. So, Jack, usingyour guess provides a bit of safety factor. Pretty goodguesstimating. :-)Clif"Education: the path from cocky ignorance to miserable uncertainty."- Mark Twain, writer and humoristMy guess was that it carrries 2/3 of the load and the rear spars only 1/3, but this is only a GUESS. It may be carrying much more of the load than that. It might be carrying =BE of the load. In Michael's case, the best thing to do is to build his plane with the cheap fittings, then load it to gross weight and subject it to maneuvers which induce 6 G's and see if he dies in the test. If he doesn't die, then it was OK. Jack Phillips NX899JP "Icarus Plummet" Raleigh, NC ________________________________________________________________________________Date: Tue, 11 Jan 2011 23:19:57 -0500
Re: Pietenpol-List: Please Check My Math (alum.wing strut)
Original Posted By: "amsafetyc(at)aol.com"
No. If you look at the chart I sent the left end is the ROOT end at the fuselage. The air flowing over the wing isalso flowing outwards on the bottom and inwards on thetop which is where tip votices come from. The air getsto the tip, flows out from underneath, rolls over the topand spirals rearwards. It's actualy doing this a little bitall the way along but nothing like at the tip. So some of the lift near and at the tip is compromised.http://www.youtube.com/watch?v=mF54SvC5ZAsClif This would also imply that there is not as much lift generated at the root either. This would mean that most of the lift would be generated in the middle 80% (paredo principle anyone.) Doug DeverIn beautiful Stow Ohio________________________________________________________________________________
No. If you look at the chart I sent the left end is the ROOT end at the fuselage. The air flowing over the wing isalso flowing outwards on the bottom and inwards on thetop which is where tip votices come from. The air getsto the tip, flows out from underneath, rolls over the topand spirals rearwards. It's actualy doing this a little bitall the way along but nothing like at the tip. So some of the lift near and at the tip is compromised.http://www.youtube.com/watch?v=mF54SvC5ZAsClif This would also imply that there is not as much lift generated at the root either. This would mean that most of the lift would be generated in the middle 80% (paredo principle anyone.) Doug DeverIn beautiful Stow Ohio________________________________________________________________________________
Re: Pietenpol-List: Adjustable Wood Wing Strut Fitting
Original Posted By: Owen Davies
Subject: Re: Pietenpol-List: Adjustable Wood Wing Strut FittingThat wasn't in the drawing just something I decided to do as a safety precaution JohnSent via DROID on Verizon Wireless-----Original message-----
Subject: Re: Pietenpol-List: Adjustable Wood Wing Strut FittingThat wasn't in the drawing just something I decided to do as a safety precaution JohnSent via DROID on Verizon Wireless-----Original message-----
Re: Pietenpol-List: Adjustable Wood Wing Strut Fitting
Original Posted By: "TOM STINEMETZE"
In response to some of the comments re the bolt pattern, embedded steel, etc. Michael said this was a preliminary drawing and I take it as such, but the concept is sound, I think.As for embedding steel in the wood strut, unless you were going to do that the entire length of the strut and use only the steel for attachment, what would be the point? Would not short lengths of steel be as prone to tear-out as the bolts? You could epoxy them in I suppose, but I've never been that impressed with the bonding strength of epoxy to steel. If you went with full length steel, what's the point to wood struts except appearance?As I stated in my original post asking bout this option, I'm making laminated struts with embedded carbon fiber running the length of the strut and additional fiber running 90 degrees to that at the ends. Of course there will be jury struts. Laminations will probably be done with West System epoxy. Does anyone find fault with that approach?KIp GardnerOn Jan 11, 2011, at 9:05 PM, Michael Perez wrote:> For whomever it was asking, here is a simple drawing of a simple > idea. There is a lot of room for change and improvement here, this > was what I came up with right off the bat.>> Michael Perez> Karetaker Aero> www.karetakeraero.com> ________________________________________________________________________________Date: Wed, 12 Jan 2011 08:27:03 -0600
In response to some of the comments re the bolt pattern, embedded steel, etc. Michael said this was a preliminary drawing and I take it as such, but the concept is sound, I think.As for embedding steel in the wood strut, unless you were going to do that the entire length of the strut and use only the steel for attachment, what would be the point? Would not short lengths of steel be as prone to tear-out as the bolts? You could epoxy them in I suppose, but I've never been that impressed with the bonding strength of epoxy to steel. If you went with full length steel, what's the point to wood struts except appearance?As I stated in my original post asking bout this option, I'm making laminated struts with embedded carbon fiber running the length of the strut and additional fiber running 90 degrees to that at the ends. Of course there will be jury struts. Laminations will probably be done with West System epoxy. Does anyone find fault with that approach?KIp GardnerOn Jan 11, 2011, at 9:05 PM, Michael Perez wrote:> For whomever it was asking, here is a simple drawing of a simple > idea. There is a lot of room for change and improvement here, this > was what I came up with right off the bat.>> Michael Perez> Karetaker Aero> www.karetakeraero.com> ________________________________________________________________________________Date: Wed, 12 Jan 2011 08:27:03 -0600
Pietenpol-List: Re: Please Check My Math (alum.wing strut)
Original Posted By: "Dennis Vetter"
Michael, Here are some photos of Kevin Purtee's Fat Bottom Girl that may help. Curt MerdanFlower Mound, TXRead this topic online here:http://forums.matronics.com/viewtopic.p ... ttachments: http://forums.matronics.com//files/100_ ... ___Subject: Pietenpol-List: Re: Please Check My Math (alum.wing strut)
Michael, Here are some photos of Kevin Purtee's Fat Bottom Girl that may help. Curt MerdanFlower Mound, TXRead this topic online here:http://forums.matronics.com/viewtopic.p ... ttachments: http://forums.matronics.com//files/100_ ... ___Subject: Pietenpol-List: Re: Please Check My Math (alum.wing strut)
Original Posted By: "Charles Campbell"
I've heard several times that on a pietenpol the front and rear spars and stutscarry about the same load.
[/b]Read this topic online here:http://forums.matronics.com/viewtopic.p ... __________
I've heard several times that on a pietenpol the front and rear spars and stutscarry about the same load.
[/b]Read this topic online here:http://forums.matronics.com/viewtopic.p ... __________> Re: Pietenpol-List: Please Check My Math (alum.wing strut)
Original Posted By:> "Jack Phillips"
> #yiv909872567 p> {margin:0;}I> actually tried to determine these numbers a while back and> put them on my web site at> http://home.comcast.net/~mmorrison123/wing_loads.html> . It's probably a goofy thing to read, but I tried> to include the effect of an elliptical lift distribution,> which is what you have with a rectangular wing.> Adjusting my numbers up for a 1200 lb airplane at 4 Gs I get> about 1760 lbs on the strut ends. Should be a little> more on the front, and a little less on the rear.> > Malcolm Morrison> http://home.comcast.net/~mmorrison123.html> > ----- Original Message -----
> #yiv909872567 p> {margin:0;}I> actually tried to determine these numbers a while back and> put them on my web site at> http://home.comcast.net/~mmorrison123/wing_loads.html> . It's probably a goofy thing to read, but I tried> to include the effect of an elliptical lift distribution,> which is what you have with a rectangular wing.> Adjusting my numbers up for a 1200 lb airplane at 4 Gs I get> about 1760 lbs on the strut ends. Should be a little> more on the front, and a little less on the rear.> > Malcolm Morrison> http://home.comcast.net/~mmorrison123.html> > ----- Original Message -----
> Re: Pietenpol-List: Please Check My Math (alum.wing strut)
Original Posted By:> "Jack Phillips"
> #yiv909872567 p > {margin:0;}I > actually tried to determine these numbers a while back and > put them on my web site at > http://home.comcast.net/~mmorrison123/wing_loads.html > . It's probably a goofy thing to read, but I tried > to include the effect of an elliptical lift distribution, > which is what you have with a rectangular wing. > Adjusting my numbers up for a 1200 lb airplane at 4 Gs I get > about 1760 lbs on the strut ends. Should be a little > more on the front, and a little less on the rear. > > Malcolm Morrison > http://home.comcast.net/~mmorrison123.html > > ----- Original Message -----
> #yiv909872567 p > {margin:0;}I > actually tried to determine these numbers a while back and > put them on my web site at > http://home.comcast.net/~mmorrison123/wing_loads.html > . It's probably a goofy thing to read, but I tried > to include the effect of an elliptical lift distribution, > which is what you have with a rectangular wing. > Adjusting my numbers up for a 1200 lb airplane at 4 Gs I get > about 1760 lbs on the strut ends. Should be a little > more on the front, and a little less on the rear. > > Malcolm Morrison > http://home.comcast.net/~mmorrison123.html > > ----- Original Message -----
RE: Pietenpol-List: Please Check My Math (alum.wing strut)
Original Posted By: owner-pietenpol-list-server(at)matronics.com [mailto:owner-pietenpol-lis
This is spanwise lift. What I was trying to approximate is the chordwise pressure distribution ' that will determine how much load is carried by the front strut compared to the rear strut. Since the center of pressure is located very close to the center of gravity (and hopefully a little bit forward of it=2C unless the plane is too tail-heavy)=2C it is loacted very close to the front spar=2C which means the front spar carries the lion=92s share of the lift load. My guess was that it carrries 2/3 of the load and the rear spars only 1/3=2C but this is only a GUESS. It may be carrying much more of the load than that. It might be carrying =BE of the load.Because you don=92t know=2C you look at what has worked in the past. If you were a real aircraft factory=2C the prudent thing to do would be to build the plane with turnbuckle fittings=2C and then intentionally load it until it breaks and see what that value was. Then divide that value by 1.5 and call that your max safe G load.In Michael=92s case=2C the best thing to do is to build his plane with the cheap fittings=2C then load it to gross weight and subject it to maneuvers which induce 6 G=92s and see if he dies in the test. If he doesn=92t die=2C then it was OK.Jack PhillipsNX899JP =93Icarus Plummet=94Raleigh=2C NC
This is spanwise lift. What I was trying to approximate is the chordwise pressure distribution ' that will determine how much load is carried by the front strut compared to the rear strut. Since the center of pressure is located very close to the center of gravity (and hopefully a little bit forward of it=2C unless the plane is too tail-heavy)=2C it is loacted very close to the front spar=2C which means the front spar carries the lion=92s share of the lift load. My guess was that it carrries 2/3 of the load and the rear spars only 1/3=2C but this is only a GUESS. It may be carrying much more of the load than that. It might be carrying =BE of the load.Because you don=92t know=2C you look at what has worked in the past. If you were a real aircraft factory=2C the prudent thing to do would be to build the plane with turnbuckle fittings=2C and then intentionally load it until it breaks and see what that value was. Then divide that value by 1.5 and call that your max safe G load.In Michael=92s case=2C the best thing to do is to build his plane with the cheap fittings=2C then load it to gross weight and subject it to maneuvers which induce 6 G=92s and see if he dies in the test. If he doesn=92t die=2C then it was OK.Jack PhillipsNX899JP =93Icarus Plummet=94Raleigh=2C NC
> Re: Pietenpol-List: Please Check My Math (alum.wing strut)
Original Posted By: > "Jack Phillips"
> #yiv909872567 p > {margin:0;}I > actually tried to determine these numbers a while back and > put them on my web site at > http://home.comcast.net/~mmorrison123/wing_loads.html > . It's probably a goofy thing to read, but I tried > to include the effect of an elliptical lift distribution, > which is what you have with a rectangular wing. > Adjusting my numbers up for a 1200 lb airplane at 4 Gs I get > about 1760 lbs on the strut ends. Should be a little > more on the front, and a little less on the rear. > > Malcolm Morrison > http://home.comcast.net/~mmorrison123.html > > ----- Original Message -----
> #yiv909872567 p > {margin:0;}I > actually tried to determine these numbers a while back and > put them on my web site at > http://home.comcast.net/~mmorrison123/wing_loads.html > . It's probably a goofy thing to read, but I tried > to include the effect of an elliptical lift distribution, > which is what you have with a rectangular wing. > Adjusting my numbers up for a 1200 lb airplane at 4 Gs I get > about 1760 lbs on the strut ends. Should be a little > more on the front, and a little less on the rear. > > Malcolm Morrison > http://home.comcast.net/~mmorrison123.html > > ----- Original Message -----
Re: Pietenpol-List: Please Check My Math (alum.wing strut)
Original Posted By: gliderx5(at)comcast.net
EF Bruhn's book "Analysis and Design of Aircraft Structures" would disagree with your statement that more lift is generated from the inboard section than the outboard sections. That would be true if the wing is tapered=2C but I'm looking at a diagram of a rectangular wing with spars of equal depth the entire length and the diagram of airload shows the same load on the tip as on the root. ----- Original Message -----
EF Bruhn's book "Analysis and Design of Aircraft Structures" would disagree with your statement that more lift is generated from the inboard section than the outboard sections. That would be true if the wing is tapered=2C but I'm looking at a diagram of a rectangular wing with spars of equal depth the entire length and the diagram of airload shows the same load on the tip as on the root. ----- Original Message -----
> Re: Pietenpol-List: Please Check My Math (alum.wing strut)
Original Posted By:> "Jack Phillips"
> #yiv909872567 p> {margin:0=3B}I > actually tried to determine these numbers a while back and> put them on my web site at> http://home.comcast.net/~mmorrison123/wing_loads.html> . It's probably a goofy thing to read=2C but I tried> to include the effect of an elliptical lift distribution=2C> which is what you have with a rectangular wing. > Adjusting my numbers up for a 1200 lb airplane at 4 Gs I get> about 1760 lbs on the strut ends. Should be a little> more on the front=2C and a little less on the rear.> > Malcolm Morrison> http://home.comcast.net/~mmorrison123.html> > ----- Original Message -----
> #yiv909872567 p> {margin:0=3B}I > actually tried to determine these numbers a while back and> put them on my web site at> http://home.comcast.net/~mmorrison123/wing_loads.html> . It's probably a goofy thing to read=2C but I tried> to include the effect of an elliptical lift distribution=2C> which is what you have with a rectangular wing. > Adjusting my numbers up for a 1200 lb airplane at 4 Gs I get> about 1760 lbs on the strut ends. Should be a little> more on the front=2C and a little less on the rear.> > Malcolm Morrison> http://home.comcast.net/~mmorrison123.html> > ----- Original Message -----
> Re: Pietenpol-List: Please Check My Math (alum.wing strut)
Original Posted By:> "Jack Phillips"
> #yiv909872567 p > {margin:0;}I > actually tried to determine these numbers a while back and > put them on my web site at > http://home.comcast.net/~mmorrison123/wing_loads.html > . It's probably a goofy thing to read, but I tried > to include the effect of an elliptical lift distribution, > which is what you have with a rectangular wing. > Adjusting my numbers up for a 1200 lb airplane at 4 Gs I get > about 1760 lbs on the strut ends. Should be a little > more on the front, and a little less on the rear. > > Malcolm Morrison > http://home.comcast.net/~mmorrison123.html > > ----- Original Message -----
> #yiv909872567 p > {margin:0;}I > actually tried to determine these numbers a while back and > put them on my web site at > http://home.comcast.net/~mmorrison123/wing_loads.html > . It's probably a goofy thing to read, but I tried > to include the effect of an elliptical lift distribution, > which is what you have with a rectangular wing. > Adjusting my numbers up for a 1200 lb airplane at 4 Gs I get > about 1760 lbs on the strut ends. Should be a little > more on the front, and a little less on the rear. > > Malcolm Morrison > http://home.comcast.net/~mmorrison123.html > > ----- Original Message -----