Today's Message Index:
----------------------
1. 12:53 AM - Re: Wing drag cables and fittings (Clif Dawson)
2. 04:42 AM - Re: Wing drag cables and fittings (Gene Rambo)
3. 06:15 AM - drag and anti-drag cables (Cuy, Michael D. (GRC-RXD0)[ASRC])
4. 06:32 AM - wing drag cables and fittings (Lawrence Williams)
5. 06:42 AM - Piet project for a song (del magsam)
6. 06:53 AM - Re: Wing drag cables and fittings (bike.mike@comcast.net)
7. 08:41 AM - Re: Wing drag cables and fittings (Michael Perez)
8. 09:08 AM - Re: Wing drag cables and fittings (Jack T. Textor)
9. 09:31 AM - for Larry Williams (Cuy, Michael D. (GRC-RXD0)[ASRC])
10. 09:54 AM - Re: Wing drag cables and fittings (Michael Perez)
11. 10:12 AM - Wing drag fitting math (Michael Perez)
12. 10:22 AM - Re: Wing drag cables and fittings (Phillips, Jack)
13. 10:57 AM - Re: Wing drag cables and fittings (Michael Perez)
14. 11:53 AM - Re: Wing drag fitting math (Phillips, Jack)
15. 12:16 PM - Re: Wing drag fitting math (Michael Perez)
16. 12:21 PM - Re: Wing drag fitting math (Michael Perez)
17. 02:02 PM - Re: Wing drag cables and fittings (Bill Church)
18. 08:11 PM - Re: Wing drag cables and fittings (amsafetyc@aol.com)
Message 1
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| Subject: | Re: Wing drag cables and fittings |
Yes, those calculations are right BUT that's not the place to
calculate from. Remember, you have a bolt holding the
fitting and strut together. That's a hole in the strap.
If the strap is 1" wide and the hole is 1/4" for instance, then
the material you have to work with is 3/4". So your strength
is now 0.75 X 0.080 X 97000.
Any failure is going to be at the weakest point. One of the
four 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 to
ease the edges by deburring them and rounding them in
the process.
Each of those four outer strut connections takes a portion
of the entire weight of the airplane. The front ones take
more than the rear. A 1200 lb aircraft in level flight has
each wing panel holding up 600 lb. Depending on the
airfoil and location of the spars, the front one is likely to
see 325 to 375 lb. That's upward force at 90=B0 to the
wing surface. The strut is at an angle so the tension on
it is quite a bit more than that upward force. Now the
wing is attached to the rest of the plane by both the outer
struts and, in our case, the inner struts. The inner ones
take 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 need
to take into account turning loads,gust loads etc.
Taking the above figures, the strap, at the hole location, has
a tensile strength of 5820 lb. The two straps then, 11640 lb.
Quite a bit stronger than needed in our application when
using 4130. You must keep in mind the Piet was originaly
designed for, and built with mild steel. In fact a lot of certified
small aircraft had mild steel in their frames.
Clif
----- Original Message -----
From: Michael Perez
To: pietenpol-list@matronics.com
Sent: Sunday, December 14, 2008 3:40 PM
Subject: RE: Pietenpol-List: Wing drag cables and fittings
So, say a wing strut fitting is 1" X 10" X .080. Would the cross
sectional area be .080? square inches? If this is right, then that
piece made of 4130 would be able to carry, (97,000 X .080) about
7,760lbs? If that fitting was doubled, as the wing strut fittings are,
(one piece each side of the spar) then the total load capability of the
fitting would be around 15,520 lbs? Or would it still be "rated" at
7,760 because that is when one side could fail and render the entire
fitting failed?
3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
3D
3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
3D
3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
3D
3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
3D
-------------------------------------------------------------------------
-----
Checked by AVG - http://www.avg.com
12/14/2008 12:28 PM
Message 2
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| Subject: | Re: Wing drag cables and fittings |
All of the math is interesting and correct, but to answer the original
post, you can rest assured that this airplane is grossly overbuilt!
Also to your original question, the drag/anti-drag wires in the wing
take very little load, especially as compared to things like lift
struts. There is no need to tighten them very much, either. once they
are drawn snug, they are doing their job. They are holding a particular
dimension and will not stretch. You can, however, do damage or add
stress by overtightening, which accomplishes nothing.
One point that may draw the wrath of the real engineers on here, but in
a setup like the lift struts, I do not think you calculate that each
strut is carrying half of the load. In a triangular setup like that,
each side carrys the entire load. I know for a fact when doing
hoisting, such as if you were lifting a steel I-beam (horizontally)with
a crane where a cable is attached to each end of the beam forming a
triangle with the crane's cable (just picture it, bear with me), each of
the two "legs" of the triangle is carrying the entire load, not half.
Gene
(ducking for cover)
Message 3
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| Subject: | drag and anti-drag cables |
This is no place to try to save weight. Use 1/8" cables, brass
turnbuckles, and per-plans 4130 steel fittings.
You get disoriented on a very hazy summer day in rapidly decreasing
visibility on your way back to the airport
and enter a rain shower, loose your horizon and find yourself passing
thru Vne and into the 100 mph zone and
you'll wish you hadn't tried to re-engineer this part of the airplane.
This is nothing to be messed with !
Mike C.
Message 4
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| Subject: | wing drag cables and fittings |
Mike-
-
I guess you've finished your leading edges and are now trammelling your win
g and making up the fittings. I know it's been alluded to before but if you
-follow the plans as far as the cables and metal gages go, you can't go w
rong. I know that Mike Cuy and others advocate lengthening some of the fitt
ings for more clearance and easier bolt insertion but I can't remember anyo
ne changing the thickness. Ask anyone about airframe failures in a plans-bu
ilt Piet and you'll get a lot of blank stares......there have been none.
-
Get on with your building......Brodhead is still 8 months away and at your
pace you should be able to fly in.
-
And pictures, WE WANT PICTURES!!!
-
Larry W.- xcg, xcmr, epp=0A=0A=0A
Message 5
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| Subject: | Piet project for a song |
if anyone is interested in a piet project that is well on its way to becoming a
beautiful flying machine. Mine is for sale for only $2000
Del
Del-New Richmond, Wi
farmerdel@rocketmail.com
Message 6
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| Subject: | Re: Wing drag cables and fittings |
Message 7
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| Subject: | Re: Wing drag cables and fittings |
Thanks guys. Good stuff! I thought for sure that the hole in the end of the
strap played a factor, I just did not think it through and do the proper m
ath. I have no intention on changing the hardware for the wing struts. I wa
s mostly curious as to the forces on the drag wires in the wings and used t
he strut fittings as an example to help me with the math.
-
It seemed to me that those cables didn't see much tension and that once the
y were in place, they would not see much in the form of other loads.
-
Mike has talked with me off list as well, so believe I have a good handle o
n it all now.
-
Lawrence, still working on the right wing. I wanted to have the actual flig
ht fittings in place temporarily to run my strings to represent the cables.
Then I can place my ribs and epoxy. So, I still just have the ribs on the
spars loose waiting for hardware to be made.
-
-
Message 8
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| Subject: | Wing drag cables and fittings |
Mike,
Good idea on the string...I did it on my second wing and it allowed all
the ribs to clear the cable. It also helps when doing the cross braces.
One more note...when you do the compression struts, make sure they don't
rise above or below your ribs.
Jack
www.textors.com
Message 9
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| Subject: | for Larry Williams |
Larry-- can you post any photos to us of your wing tubing setup before
cover ?
You know you could have capped that baby off on either end, welded in a
drain fitting and used it
as your fuel tank leaving your wing center section free for baggage
space.
I'm very proud of you for coming out of the closet about how you did
your wing----you were holding back
on us man !
Mike C.
<<Bhead05LarryWilliamsPanel.jpg>>
Message 10
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| Subject: | Wing drag cables and fittings |
Good note on the compression struts Jack! Thanks.
Message 11
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| Subject: | Wing drag fitting math |
Group, I have done some math on these fittings. I am sure some of you may h
ave done it before, but I want to show what I did and see what you think.
-
>From what I found, 4130 has a tensile strength of 79,000 PSI. T6061-T6 alum
inum is rated at 45,000.
-
The smallest cross sectional area fitting for the wing drag cables is 5/8"
wide by .080. (the thickness of the 4130 steel.)
-
Here is what I came up with using the 3/16" hole as in the prints:
-
.625 - .1875 = .4375 (5/8" wide minus the 3/16" hole)
.4375 X .080 = .035 (cross sectional area of that 5/8" fitting with a 3/1
6" hole)
.035 X 79,000 = 2,765 pounds.
-
-
Next I did that same fitting made out of .125 aluminum.
.625 - .1875 = .4375
.4375 X .125 = .0546875 (cross sectional area of the 5/8" aluminum with t
he same hole)
.0546875 X 45,000 = 2,460.938 lbs.
-
The 1/8" galv. 7 X 19 cable at A.S. is rated at 2,000 lbs.
-
If I did the math right with the right formulas, would aluminum be a good s
ubstitute for these fittings? The other drag wire fittings are 3/4" wide. I
f 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 t
o figure out THOSE numbers?- Thanks all.
Message 12
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| Subject: | Wing drag cables and fittings |
Jack's 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" plywood,
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, and to keep them
from slipping sideways due to shock loads. I don't 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 Phillips
NX899JP
Raleigh, NC
From: owner-pietenpol-list-server@matronics.com
[mailto:owner-pietenpol-list-server@matronics.com] On Behalf Of Michael
Perez
Sent: Monday, December 15, 2008 12:54 PM
Subject: RE: Pietenpol-List: Wing drag cables and fittings
Good note on the compression struts Jack! Thanks.
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orsk - Portuguese
Message 13
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| Subject: | Wing drag cables and fittings |
Thanks Jack. I had seen that type of setup before and nay use it as well.
Message 14
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| Subject: | Wing drag fitting math |
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" aluminum, with a #10 bolt (AN3) the edge distance would be 3/8".
After subtracting half the bolt diameter (since that distance was
measured to the centerline of the bolt), you are left with .375" -
.188"/2, or .281". This gives a shear area of .281" x .125" (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't 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's 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" thick and 13'
(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't 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's comment to not make the cable any tighter than necessary is good
advice. It doesn't 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't want to use aluminum for your wing
fittings. It also should convince you that you really want 1/8" cable
here and in any high load areas, and use 3/32" for areas of lesser load.
3/32" cable has less than half the strength of 1/8".
Jack Phillips
NX899JP
_____
From: owner-pietenpol-list-server@matronics.com
[mailto:owner-pietenpol-list-server@matronics.com] On Behalf Of Michael
Perez
Sent: Monday, December 15, 2008 1:12 PM
Subject: Pietenpol-List: Wing drag fitting math
Group, I have done some math on these fittings. I am sure some of you
may have done it before, but I want to show what I did and see what you
think.
>From what I found, 4130 has a tensile strength of 79,000 PSI. T6061-T6
aluminum is rated at 45,000.
The smallest cross sectional area fitting for the wing drag cables is
5/8" wide by .080. (the thickness of the 4130 steel.)
Here is what I came up with using the 3/16" hole as in the prints:
.625 - .1875 = .4375 (5/8" wide minus the 3/16" hole)
.4375 X .080 = .035 (cross sectional area of that 5/8" fitting with a
3/16" hole)
.035 X 79,000 = 2,765 pounds.
Next I did that same fitting made out of .125 aluminum.
.625 - .1875 = .4375
.4375 X .125 = .0546875 (cross sectional area of the 5/8" aluminum with
the same hole)
.0546875 X 45,000 = 2,460.938 lbs.
The 1/8" galv. 7 X 19 cable at A.S. is rated at 2,000 lbs.
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.
_________________________________________________
This message is for the designated recipient only and may contain privilege
d, proprietary
or otherwise private information. If you have received it in error, please
notify the sender
immediately and delete the original. Any other use of the email by you is p
rohibited.
Dansk - Deutsch - Espanol - Francais - Italiano - Japanese - Nederlands - N
orsk - Portuguese
Message 15
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| Subject: | Wing drag fitting math |
Thanks Jack! That was what I was looking for. I had asked in one of my earl
ier posts what the load/tension was on the drag cables as installed. I also
knew of flight loads, but no one responded with any numbers.
-
I appreciate your time very much. I enjoy learning these type of things as
they apply in other areas other then aviation.
-
--- On Mon, 12/15/08, Phillips, Jack <Jack.Phillips@cardinalhealth.com> wro
te:
From: Phillips, Jack <Jack.Phillips@cardinalhealth.com>
Subject: RE: Pietenpol-List: Wing drag fitting math
If I did the math right with the right formulas, would aluminum be a good s
ubstitute for these fittings? The other drag wire fittings are 3/4" wide. I
f 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 t
o 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 th
e bolt and the length of whatever edge distance the fitting was made with.
- Typically, the edge distance should be twice the hole diameter (edge di
stance being measured from the edge of the part to the centerline of the ho
le) 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 ha
lf 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 give
s 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 str
ength).- This means that the load required for the fitting to fail in she
ar is 22,500 x .035 sq in., or 790 lbs.- Note that this also does not app
ly 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 app
lied to any engineering loading
-
It is generally good design to make the fittings and hardware at least as s
trong as the cable so that if the plane is overloaded, the cable will stret
ch 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 w
ill see are the initial installation loads.- The primary purpose of the d
rag 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=AxPx
Cd, 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 ge
nerally wind pressure is taken as 0.00256 V2, where V is the airspeed in mp
h.- 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 (ma
kes the calculations easier).- If you just take the frontal area of the w
ing 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 f
orce 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 p
lanes 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 p
retty sharp ' about 19 degrees.- This means that for the cables to resi
st 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 wh
atever tensile load is present in the cable at rest.- This is why Gene Ra
mbo=92s comment to not make the cable any tighter than necessary is good ad
vice.- It doesn=92t take much preloading to get the cable up to several h
undred 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 loa
d 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 br
acing 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 Phillips
NX899JP
-
From: owner-pietenpol-list-server@matronics.com [mailto:owner-pietenpol-lis
t-server@matronics.com] On Behalf Of Michael Perez
Sent: Monday, December 15, 2008 1:12 PM
Subject: Pietenpol-List: Wing drag fitting math
-
Group, I have done some math on these fittings. I am sure some of you may h
ave done it before, but I want to show what I did and see what you think.
-
>From what I found, 4130 has a tensile strength of 79,000 PSI. T6061-T6 alum
inum is rated at 45,000.
-
The smallest cross sectional area fitting for the wing drag cables is 5/8"
wide by .080. (the thickness of the 4130 steel.)
-
Here is what I came up with using the 3/16" hole as in the prints:
-
.625 - .1875 = .4375 (5/8" wide minus the 3/16" hole)
.4375 X .080 = .035 (cross sectional area of that 5/8" fitting with a 3/1
6" hole)
.035 X 79,000 = 2,765 pounds.
-
-
Next I did that same fitting made out of .125 aluminum.
.625 - .1875 = .4375
.4375 X .125 = .0546875 (cross sectional area of the 5/8" aluminum with t
he same hole)
.0546875 X 45,000 = 2,460.938 lbs.
-
The 1/8" galv. 7 X 19 cable at A.S. is rated at 2,000 lbs.
-
If I did the math right with the right formulas, would aluminum be a good s
ubstitute for these fittings? The other drag wire fittings are 3/4" wide. I
f 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 t
o figure out THOSE numbers?- Thanks all.
-
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| Subject: | Wing drag fitting math |
Jack, I just read your post again, what good stuff. I believe you answered
all my questions.- Thanks.
Message 17
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| Subject: | Wing drag cables and fittings |
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 necessary
knowledge to assume that they could safely reduce the size of almost any (or
all) component(s) of the plane. This simply is not the case. Since the plans
were drawn in an age where mild steel was the norm for aircraft
construction, and today considerably stronger 4130 is the norm, there are
some areas where there is potential to safely modify certain components
(again, assuming that one has the requisite knowledge to do the necessary
calculations). But, in general, if built to the plans, everything should
work 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 at
the upper end of the speed range), the draggy old Piet wing may very well
witness significant loads. The cables wouldn't be where they are, and sized
as they are for no reason. Your point about not overtightening the wires is
a very good one. The cables are there to carry loads imposed on them by the
drag 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". Your
example of a steel beam being lifted with a cable attached to each end is an
interesting one. Let's assume the beam weighs 1000 pounds. In an ideal
world, 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 zero
horizontal component to the load. Since all that leaves is the vertical
load, each cable takes half, or 500 pounds. Unfortunately, most of us live
in the real world, and we cannot use infinitely long cables. In order to
make use of limited space, the cables are shortened, resulting in a
triangular set-up, as you mentioned. When the arrangement changes from
vertical cables to angled cables, horizontal loads are imposed. The vertical
component remains at 1000 pounds (due to gravity), but the horizontal
component increases with each degree that the cable moves off of vertical.
These horizontal forces are ADDED to the vertical forces, to give a
resultant force, which acts along the length of the cable. Typically, in
practice, the angle formed between the beam and the cable is not less than
30 degrees. At 30 degrees, the resultant force acting on the cable is
exactly twice the vertical component. For this reason, each of the cables
used to lift a 1000 pound beam needs to be rated for 1000 pounds, since each
end will actually be loaded at 1000 pounds. If the angle is less than 30
degrees, the load goes up even higher. For instance, if the angle is only 20
degrees, the resultant load will be almost 1500 pounds. If the angle is only
10 degrees, the resultant load is almost 3000 pounds. These calculations are
not complicated, just using basic trigonometry (sine).
So, in practical terms, if you're lifting a beam with cables arranged at 30
degrees, each of the cables does need to be strong enough to carry the full
weight of the beam, since that IS the (resultant) load each cable will be
loaded 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 of
the beam (in this case).
I've attached a simple sketch showing the three cases mentoned above (30, 20
and 10 degrees). By carefully drawing accurate triangles to scale, it is
possible to calculate the resultant loads without dredging up unpleasant
(for many) memories of high school trigonometry.
Bill C.
_____
From: owner-pietenpol-list-server@matronics.com
[mailto:owner-pietenpol-list-server@matronics.com] On Behalf Of Gene Rambo
Sent: Monday, December 15, 2008 7:42 AM
Subject: Re: Pietenpol-List: Wing drag cables and fittings
All of the math is interesting and correct, but to answer the original post,
you can rest assured that this airplane is grossly overbuilt! Also to your
original question, the drag/anti-drag wires in the wing take very little
load, especially as compared to things like lift struts. There is no need
to tighten them very much, either. once they are drawn snug, they are doing
their job. They are holding a particular dimension and will not stretch.
You can, however, do damage or add stress by overtightening, which
accomplishes nothing.
One point that may draw the wrath of the real engineers on here, but in a
setup like the lift struts, I do not think you calculate that each strut is
carrying half of the load. In a triangular setup like that, each side
carrys the entire load. I know for a fact when doing hoisting, such as if
you were lifting a steel I-beam (horizontally)with a crane where a cable is
attached to each end of the beam forming a triangle with the crane's cable
(just picture it, bear with me), each of the two "legs" of the triangle is
carrying the entire load, not half.
Gene
(ducking for cover)
Message 18
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| Subject: | Re: Wing drag cables and fittings |
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