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Frame Stress From High Winds


kairusan

Rev Arcana Poll Version 1.0 alpha  

42 members have voted

  1. 1. Which parts of the Rev 1.5's frame are most apt to be riven asunder by the raw elemental fury of high winds?

    • The vertical spars.
      24
    • The leading edge spars.
      11
    • liek who carez LOL
      8


Question

I'm curious as to what parts of a 1.5's frame — the verticals or the leading edge — receive the greatest amounts of stress from high winds. I've done some searches on the board and have found some conflicting opinions; most people seem to say that the leading edge takes most of the stress and the verticals absorb very little, but there are a few contrary opinions (including one from the Great Barresi himself, if I am not mistaken). Perhaps a bit of informal internet democracy (with commentary) is in order, so that this eternally vexing question might at last be resolved once and for all? Perhaps not? You decide. :blue-cool:

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Only broken one rod so far - knock on wood, assembling LE and didn't have outer rod seated fully, gust came up and cracked end of rod! ::kid_cussing::My Bad :censored: Still have 4 out of 5 original rods from my Rev 1!! Tough little buggers!!!:)

I wonder if anyone can translate the following into something that relates to tubular rods <grins>

Felix

"Imagine we have a solid circular rod, and consider the forces in the

middle as we try to bend it. I'm going to draw some cross sectional

diagrams.

Seen from the side:

y coordinate

| axes plane of interest

|___x v

<=== ===> F

---------------------------------------------

C _ ^ stretched : : _

/ \ |2r : deviation, / \ C

\> | : exaggerated </

v squashed *

---------------------------------------------

Seen end-on:

+----> expansion

y e

| ,-------. | /

|___z / \ | /

| | |/ graph of

| | / expansion

| | /| vs y

\ / / |

`-------' / |

So we have a rod of radius r. We apply a bending couple C. This will

cause it to bend slightly, into (by symmetry) a section of a very

large radius hoop. We will neglect the 2nd-order effects of squeezing

of the rod in the y direction and expansion in the z direction.

>From the geometry, we can see that the expansion or contraction

distance is proportional to y. At the middle (y=0) the rod neither

stretches or compresses; at the top edge (y=r) it is maximally

extended. Ie e(y) = d * y where e is the proportion by which the

material is extended and d is a value indicates how tight a bend we

get.

While the tube is still in its elastic region (ie, it hasn't started

to bend permanently ie fail), the force is proportional to the

extension. Since the force varies across the cross-section of the

rod, we want the force at each point - ie, the pressure. The

geometrical symmetry means the force varies only according to y and

not x or z.

Ie, P(y) = E * e(y) Hooke's law, where E is a material

property (Young's modulus).

Consider the component of the couple at the plane of interest,

provided by the horizontal strip of the plane of interest at the

vertical position y:

dC(y) = dF(y) * y

where dC is the sliver of the couple and dF is the sliver of the

normal force. But force is pressure times area:

dF(y) = P(y) * w(y) * dy

where w is the horizontal width across (ie, in the z direction) the bar

at the height y. Since the bar is circular in cross-section,

w(y) = sqrt( r^2 - y^2 )

So substituting all that in,

dC(y) = ( E * ( d * y ) ) * y * sqrt(r^2 - y^2) * dy

= E * d * y^2 * sqrt(r^2 - y^2) * dy

(NB I have neglected to think too hard about the overall sign of

this couple.)

Ie to calculate C we have to integrate the contribution of the P at

the various y:

r

/\

C = | E * d * y^2 * sqrt(r^2 - y^2) * dy

\/

y = -r

I didn't fancy remembering how to integrate that so I cheated and used

Wolfram Mathematica online integrator[1] which tells me:

C =

- E * d * 1/8 *

[ -1 y ] +r

[ y * sqrt(r^2-y^2) * (2*y^2 - r^2) + r^4 tan ------------- ]

[ sqrt(r^2-y^2) ] y=-r

The LH term is zero at both limits because sqrt(r^2-y^2) is zero at

both limits. So we are left with the RH term. This is slightly

awkward the way Wolfram has integrated it (using the 1-argument atan

rather than the computer person's 2-argument atan) because the

denominator is zero, but atan is actually well-defined if we take the

appropriate limit. So:

[ -1 y ] +r

C = -1/8 * E * d * [ lim l^4 * tan ------------- ]

[ y=0..->l sqrt(r^2-y^2) ] l=-r

The sqrt() denominator approaches 0 from the positive side in each

case. l^4 is always positive. The numerator is positive at the upper

limit and negative on the lower, so:

-1 -1

C = -1/8 * E * d * r^4 * ( lim tan f - lim tan f )

f->inf f->-inf

= -1/8 * E * d * r^4 * ( pi/2 - -pi/2 )

= -pi/8 * E * d * r^4

The minus sign is simply due to carelessness about the overall sign

and can be disregarded. Recollect that E is a material parameter and

d is a geometrical meausre of how much we've bent the rod. Let K =

pi/8 * E, and dropping the minus sign, we have:

C = K * d * r^4

So for any given material, and amount of bending, the couple

(informally, force) required to bend the rod by that amount is

proportional to the 4th power of the rod's radius. The rod's

stiffness against bending goes as the 4th power of the radius.

This is the Luggage's 4th power.

But if we are interested in the _failure strength_ we want to know

when the rod will fail. Failure occurs when the expansion of the rod

at any point exceeds the material's elastic limit, ie where:

r

max e(y) >= e

y=-r lim

As soon as that is reached anywhere that bit of the rod will start to

yield, and the other parts will need to take up the strain and won't

be able to do so, and it will buckle (this is not true for all

structures but it is for this one). e_lim is a material property too.

Since e(y) = d * y obviously the maximum e is at r. (It will

probably fail at the top, in tension, but if it fails in compression

that's -r, which will make no difference to the analysis.)

So we have, at failure:

d * r = e

lim

What we want to know is how C_lim (the couple at which the rod fails)

depends on r.

Above we had:

C = K * d * r^4

Substituing in to eliminate d, we get:

e

lim

C = K * ---- * r^4

lim r

or

C = K * e * r^3

lim lim

So the rod's _strength_ against bending is only proportional to the

third power of its thickness, not the fourth.

The discrepancy between third power for strength and the fourth power

for stiffness is because a thicker rod bends less before it fails.

[1] http://integrals.wolfram.com/index.jsp?expr=x^2+*+Sqrt[r^2+-+x^2]"

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Can anyone translate that into something a common "joe" can get his mind around???kid_devlish.gif I think I gave up after the first paragraph!!!smiley-devil.gif

I think that I recollect, sometime ago, that I read where John said "the verticals will always go first". John, please forgive me if I have misquoted, and please set us straight.....................confused_1.gifconfused_1.gif

nick

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No no; Felix is serious. A solid rod is just a tube with zero inside diameter.

Of course, Felix, you have to consider the fact that wrapped carbon fiber tubes have different strengths in the lateral and longitudinal directions.

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No no; Felix is serious. A solid rod is just a tube with zero inside diameter.

Of course, Felix, you have to consider the fact that wrapped carbon fiber tubes have different strengths in the lateral and longitudinal directions.

Uh! Quite...

Rods break, very occasionally. Loose ferrules seem to be the main danger in the leading edge. It is worth checking them as a matter of habit, especially with team kites.

Felix

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laugh.gif I'm always willing to escalate this sort of thing.

The Green Race rods seemed to spring back more quickly after stress/gusts in the JB Pro Vented today...

I still do not have a weight comparison between Green and Standard Race!

Felix

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...I still do not have a weight comparison between Green and Standard Race!

Felix

I have a nice, accurate digital scale that covers this range of weights. Now, if anyone would like to send me a set of each, I would certainly be glad to weigh them and let you know... laugh.gif

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I have a nice, accurate digital scale that covers this range of weights. Now, if anyone would like to send me a set of each, I would certainly be glad to weigh them and let you know... laugh.gif

And you'll return them in due time??!!kid_devlish.gif

<grins>

I guess someone will deal with this in due course...

Felix

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Couldn't Revolution supply you with the actual weight, of the different rod types? Weighing one individual set of rods, really isn't very accurate, as I am sure there is a bit of variation, within individual rods, as well as from rod to rod. The best case scenario, would be for someone, with access to lots and lots of rods, to weight at least a 25, or 50, or even 100 rods, all at one time, and then do the math to come up with a realistic average weight per rod. Then you need to do the same thing with the ferrules, which are used in the center sections, so that two of those weights, could be added to 5 of the rod weights, to come up with an accurate weight for a set (whatever set you may be interested in)..............

Seems like some information, that the quality control area, at Revolution Kites, should have privy to, (or Ben), if that isn't proprietary information...........

Just a suggestion.............blue_confused.gif

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Couldn't Revolution supply you with the actual weight,

<snip>

Seems like some information, that the quality control area, at Revolution Kites, should have privy to, (or Ben), if that isn't proprietary information...........

Just a suggestion.............blue_confused.gif

My question was 'rhetorical' as I have every confidence in the company providing kit with good reason.

The 'spring time' of the standard and green Race Rods seems to be a case in point. Good job Ben... I like the 'snap back' of the green rods. especially as compared to the bamboo equivalent...

The actual physics can, EDIT <I presume>, 'obviously' be substantiated.

Felix

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{ how to get a realistic weight for a set of rods }

Seems like some information, that the quality control area, at Revolution Kites, should have privy to, (or Ben), if that isn't proprietary information...........

I will offer a guess that this isn't something that has already been done. It isn't really required for QA (although testing for a rod with the weight really off would be a good easy test for a number of faults). This sort of information usually only gets collected when someone gets curious. Once the information does get collected, very often someone will discover something useful to do with it.

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I will offer a guess that this isn't something that has already been done. It isn't really required for QA (although testing for a rod with the weight really off would be a good easy test for a number of faults). This sort of information usually only gets collected when someone gets curious. Once the information does get collected, very often someone will discover something useful to do with it.

Ummmmm ...... I hadn't really thought that far into the future. ....... Really innocent.gif

After all, it might take years to find the scale. laugh.gif

Aha! So the scales may not be immediately to hand? <grins>.

Never mind, I have not got spare sets of rods to send off for testing so I will 'stick' to anecdotal responses for the time being.

If necessary we will re-build the 'bamboo' framed kite that flew in Vienna all those years ago. I suspect that the original kite is still hanging on a wall in that fair city...

Felix

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I'd like to buy a cyber-drink for the 7 folks who answered "liek who carez LOL" on the poll attached to this topic. :kid_smartass:

Heck; I'll buy a real one for the first 7 who corner me at some kite fest (and remind me that I said so). Throw one in for you, too, John.

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I'd like to buy a cyber-drink for the 7 folks who answered "liek who carez LOL" on the poll attached to this topic. kid_smartass.gif

Heck John, You ought to be able to find out that information, if it's been generated, and again, if it isn't proprietary information...........blue_wink.gif

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Heck; I'll buy a real one for the first 7 who corner me at some kite fest (and remind me that I said so). Throw one in for you, too, John.

Hey, how about me ? Just kidding Pete. Why does "Felix" want the weights of the race rods? Is it just because he sees a difference in their performance, or is it a weight thing? confused_1.gif

smile.gifjust for you smile.gifand just kidding you smile.gif

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Was it me trying to find everything I could about green vs regular Race Rods, I think I'd grab a set of each in either hand and "bounce" them. Then I'd swap hands and "bounce" them again. If I couldn't tell the difference, I would judge them to be "close enough". (Bouncing and swapping lets you compare weights fairly accurately. For remote comparison, you need accurate scales.)

Then I'd arrange a couple of blocks about rod-length minus an inch apart, and lay a rod across and hang a weight from the middle with a loop of kiteline. Lay a straight-edge across the rod and measure the deflection. Then do the same with the other race rod. Compare the deflections. If you want to be able to compare with other people remotely, the distance between the blocks and the amount of weight should be specified.

I'm willing to do what I can locally, but as I said, no race rods here. I could weigh and deflect 3-wrap and 4-wrap rods, as well as the T300 Hi-Modulus rods from my old Rev I.

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Was it me trying to find everything I could about green vs regular Race Rods, I think I'd grab a set of each in either hand and "bounce" them. Then I'd swap hands and "bounce" them again. If I couldn't tell the difference, I would judge them to be "close enough". (Bouncing and swapping lets you compare weights fairly accurately. For remote comparison, you need accurate scales.)

Then I'd arrange a couple of blocks about rod-length minus an inch apart, and lay a rod across and hang a weight from the middle with a loop of kiteline. Lay a straight-edge across the rod and measure the deflection. Then do the same with the other race rod. Compare the deflections. If you want to be able to compare with other people remotely, the distance between the blocks and the amount of weight should be specified.

I'm willing to do what I can locally, but as I said, no race rods here. I could weigh and deflect 3-wrap and 4-wrap rods, as well as the T300 Hi-Modulus rods from my old Rev I.

Ok, gottcha! I hadn't read back far enough.................. laugh.gif

Oh, I only got regular race rods (no green ones), and then of course, 2-wraps, 3-wraps, & 4-wraps, so I can't help. smile.gif

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