shrewdPanther46 wrote:WhatScience? wrote:Balsa Man wrote:
What do you mean, "better"?
Which would be more efficient?? Would you say the weight matters more or the additional surface area
depends on ur tower as stated previously
theres no such thing as "better"
Let me see if I can bring some clarity to this discussion. There
is such a thing as “better” in the sense that there is some combination of leg wood and bracing wood that gets you to the lightest possible tower that will carry full load, or very close to it.
The design problem is figuring out what that combination is. So, weight matters, ultimately - you want to get to the lightest tower, you can that can carry close to full load. “Surface area” has… nothing to do with anything. Cross section does have something to do with it.
It is all about buckling strength. The legs (actually the leg segments- by which I mean the upper, chimney leg segments, and the lower, base segments need to have enough buckling strength to carry the forces they come under when the tower is loaded. Let’s call that our “design strength.” As discussed previously, the forces on the chimney legs are less than the forces on the base legs, so you have to look at/analyze the upper and lower segments separately; they have different design strengths.
There is, at the heart of this design problem, a tradeoff going on that has to be figured out. With really stiff/heavy legs, to get to our ‘design strength’ you can use less bracing- that is, you can use a wider bracing interval, which means less bracing wood needed. With really light/less stiff- floppy legs, you have to go to a much tighter/closer bracing interval, which means more bracing wood. The question is, what combination of leg weight and bracing weight (and glue weight) gets you to the lightest tower weight that just gets you to the design strengths.
Yours is the second question in the last 2-3 weeks that goes to a very important factor in solving this design problem- how does the size - the cross section - of the legs affect buckling strength (and weight)? Before going into that, the short answer to your question- 3/32” legs at 0.6gr/36” will get you to a lighter tower weight that will carry near full load than 1/16” at 0.3gr/36”, and 1/8” legs at around 1.2gr/36” will get you to a lighter tower weight (that will carry full load) than the 3/32” legs will. Thinking that smaller cross section legs will get you higher tower efficiency is going in the wrong direction. As I’ve said before, as you go from small cross section to larger cross section, at the same density, the buckling strength goes up much faster than the weight goes up. Putting actual numbers to this… reality takes some fairly serious number crunching, though. Its not complicated math, You just need to be careful.
While I’ve gone over all the information that follows in previous posts, I’m going to re-cover some of this ground. A few days ago I was discussing with one of the students I’m working with how buckling works, and how we’re using that understanding to “engineer” and design our towers. We’d gone over all this before, and I thought he understood, but it became apparent that some really fundamental aspects had… just gone by over the top of his head. He was wondering the same thing you were- why don’t we just use smaller, lighter legs to get to a lighter tower. After circling back, and explaining those fundamental aspects, the light bulb went off; what we were doing, and how we were doing it, to design/engineer the towers, made sense, and the pieces fit together. The conversation went something like this:
Remember, at the outset of this discussion, that one of the very important aspects of what “engineering” is, and is all about, is developing and using approximations that “work” to design and create… something physical, for a given use. To “work”, the physical something has to stand up to the forces it will experience in its intended use. These approximations are typically simplifications of the full underlying science (all the theoretical physics and mathematics), and often/typically require some simplifying assumptions. If you google “euler buckling” or “Euler buckling equation”, you’ll find this equation in various forms, and a LOT of seriously advanced, complicated math, much of it well beyond anything most high school students are exposed to (and beyond what I can understand and use).
F = pi squared x E x I divides by (K x L) squared
You need to understand/remember in using a formula like this to “do engineering” – you don’t just plug numbers into this equation for the various factors, and get an answer/the answer. You use/apply the relationships the equation shows/defines. It is the relationships that matter. The formula is a short-hand summary statement in mathematical form saying (in this case) when you want/need to understand how a long, thin column, made of a homogeneous, elastic material behaves when put under an axial compression force behaves, here are the factors that determine the behavior, that you need to take into account, and here are the relationships between those determining factors- how they work and play together- how the BS of a stick/piece of wood changes as you change any of the factors that determine BS. Remember, wood is not a homogeneous material.
The formula was first published in 1757 by the Swiss mathematician Leonard Euler. It came from his (mathematical) study of how long, thin columns deflected (and hence failed) when a load/force was applied at the ends, along the long axis of the column
In this equation"
F = Euler's critical load (with longitudinal/”axial” compression load on column) = the buckling strength; “BS”). It is the load/force that a column, a tower leg, will start to bow/deflect. That bowing/buckling will happen in the middle of the column.
What are the factors this is telling us that determine the BS?
They are:
E = the modulus of elasticity (aka Young’s modulus) of column material. This is the inherent stiffness of a piece of wood; it doesn’t depend on, it doesn’t have a relationship to size/cross section. It does have a linear relationship to the density. As discussed in previous posts, as wood density increases, E increases. On average, if you double the density (increase it by a factor 2), E goes up by a factor of about 2.25, but there is significant “variation around the mean”- two sticks, same size and same weight can have significantly different E values, which means significantly different BSs. This is what I was talking about when I noted wood is not a homogeneous material.
I = minimum area moment of inertia (sometimes referred to as the 2nd moment of inertia) of the cross section of the column. For a square cross section leg, where the length of the sides are x, I = (x^4)/12. (that’s x to the fourth power, divided by 12
L= unsupported length of column, This can be the length of a stick you’re testing the buckling strength of; it can the length of a leg segment (i.e., chimney section leg segment, base section leg segment), or it can be the length of a ‘braced interval’ – the length between braced points along the leg, the length of the short ‘stacked’ columns you create when a leg is braced.
K = column effective length factor. This comes from the “end conditions” of the column, or section of column you’re evaluating. When you do ‘single finger push-down’ (SFPD) testing of a stick to measure its BS, the end conditions approach what is called ‘pinned-pinned’ end conditions. If you brace this stick using a ladders and Xs bracing approach, both ends of the shorter braced segment approach what is called ‘fixed-fixed’ end conditions. The effective length factor for this situation turns out to be about 2.3. So, if you measure SFPD BS of a stick at 10 grams, you use 23gr (10gr x 2.3) as the value for doing inverse square calculations to see what bracing interval you need to use to have the leg carry your design load.
The effective length factor for all Xs is significantly less; about 0.55. Note, as I’ve explained before, that this factor was derived from - was back-calculated from - detailed wood specifications I was provided from a couple of State winning towers. While I understand the theoretical basis for getting to the 2.3 effective length factor to convert pinned-pinned end conditions to fixed-fixed, I don’t yet have that understanding for converting pinned-pinned to… whatever the end conditions are when braced with all Xs. What matters for us is that both effective length factors “work”; when you design using them, the tower, if carefully constructed, holds the weight you’ve calculated. In the all Xs case, when I say it works, I’m saying that using 1/16” x 1/32” Xs works and the effective length factor 0f 0.55 works. I don’t know how different the factor might be if using different size/cross section X strips.
So, to look at a question like you pose, you need to, how do I say this… isolate on the relationships between variables. In doing the inverse square calculations, we isolate on the length factor, and apply the inverse square relationship between length and buckling strength. In looking at the effect of cross section on buckling strength (for square legs), first, we need specific information on size. We then can look at the relationship of I to the cross sectional width. So, let’s look at 4 cross sections, 1/16”, 3/32”, 1/8”, and 5/32”.
1/16” = 0.0625”. That, to the 4th power/12 = 0.0000012717
3/32” = 0.09375”. That, to the 4th power/12 = 0.000006437
1/8” = 0.125”. That, to the 4th power/12 = 0.00002035
5/32” = 0.15625”. That, to the 4th power/12 = 0.000103
So, if looking at two sticks, we assume they have the same E (and simplifying things, assume that same E means same density) and we assume they have the same length, the relationship between the BSs of the two sticks will be the relationship between I values:
The BS of the 3/32” stick will be 5.06 times that of the 1/16” stick. The BS of the 1/8” stick will be 3.1605 x that of the 3/32 stick, and 16 x that of the 1/16” stick. The BS of the 5/32 will be 2.44 x that of the 1/8 stick (and 7.716 x that of the 3/32 stick, and 39.06 x that of the 1/16” stick).
Then, we need to look at the relationship between the weights of the sticks at these sizes, assuming they are all of the same density; have the same “E”. The weight of the 3/32” stick will be 9/4 (=2.25x) that of the 1/16” stick (it will take 4 1/32” sticks to make a 1/16” stick; it will take 9 1/32” sticks to make a 3/32” stick); 9/4 = 2.25. The 1/8” stick will be 1.78 x that of the 3/32 stick (16/9); the weight of the 5/32 will be 1.56 x that of the 1/8 stick (25/16) (and 2.78 x that of the 3/32 stick).
In all cases, looking at the relative increase in both weight and BS, as you increase stick size, the BS goes up significantly faster than the stick weight goes up.Here’s a little table looking at these relationships
size Relative WGT Relative I
1/16” 1 1
3/32” 2.25 5.06
1/8” 4 16.00
5/32” 6.25 39.06
So, finally, to your specific example, let’s put some numbers to it:
Starting with a 1/16” stick. Based on the density number you gave (0.3), I assume you are talking about a 1/16” x 36” stick weighing 0.3 grams. So, a 3/32” x 36” stick of the same density (and we’re assuming same E) would have a weight of 2.25 x 0.3, which would = 0.675gr. That means your 3/32” stick at a weight of 0.6gr would have a density about 11% lower than that of the 1/16” stick at 0.3gr. At the same density, the 3/32” stick would have a BS 5.06 x that of the 1/16” stick. If you test the BS of a few 1/16” sticks weighing close to 0.3gr, you’ll see readings around 1.8gr. If you test a few 3/32” sticks weighing around 0.68gr, you’ll see BS measurements around 9.1gr (5.06 x 1.8). Testing 3/32” sticks weighing 0.6gr, you’ll see a BS a bit lower…8-ish grams.
To keep things a bit less complicated, let’s look at sticks of the same density, at 1/16” and 3/32”, and let’s throw 1/8” sticks into the analysis. And let’s look, for this analysis, at the base segment for a B tower (which would be a leg segment length of 27.05cm.
The 1/16” 36” sticks weigh 0.3gr, and have a BS of 1.8 gr. The 3/32” sticks weigh 0.68gr and have a BS of 9.1 gr, and the 1/8” sticks weigh1.2gr, and have a BS of about 28.8gr.
So, which is “better”/”more efficient”?? You can’t really say anything just from just this information, because it depends on the bracing interval you have to use to get to the braced leg BS up to design strength. What we need to figure out for these three leg size options is which one gives us lower leg weight+bracing weight+glue weight.
First, let’s figure out the bracing interval needed to make each of these stick sizes/weights work. To do this, I’ve used the inverse square table I posted a link to, a couple pages back, and have changed stick 36” BS to look at the BSs of our 1/16” and 3/32” sticks that we just came up with.
For our “design strength” for this analysis, we’ll use 4499gr, which is applying a 10% safety factor to the 4090gr force a base segment leg in B tower meeting the 29cm circle bonus will see at a 15kg tower load.
If you look at the inverse square table, you’ll see that for 1/8”, with a 36” SFPD BS at 28.8gr, using a 1/5 bracing interval (using all Xs, with the Xs at 1/32 x 1/16) we get a braced leg strength very close to our 4499gr design strength (BS at 28.9 gets us 4524 (cell AG38).
Looking at the 1/16” stick, to get over 4499gr braced leg strength, we need to go to a bracing interval of 1/20, and for the 3/32” stick, we need to go to a bracing interval of 1/9.
So, moving one more step toward an answer, what does that mean in terms of tower segment weight?
With our 1/16” leg wood, we are looking at a total leg weight of 0.35gr
With our 3/32” leg wood, we are looking at a total leg weight of 0.71gr
With our 1/8” leg wood, we are looking at a total leg weight of 1.42gr
Having worked out the total length of bracing pieces for 1/5 interval (579cm), roughly calculated, the total length of bracing pieces for a 1/9 bracing interval is about 1037cm. For the case of a 1/20 bracing interval, we’re looking at almost 4 times that for a 1/5 interval, so about 2300cm.
So, now we can estimate bracing weight. Using 7.5gr 3” x 36” x 1/32” sheet:
For the 1/16” legs, we’ll need 3.92gr of Xs
For the 3/32” legs, we’ll need 1.77gr of Xs
For the 1/8” legs, we’ll need 0.99gr of Xs
Then for the glue. Using 0.01gr/joint:
For the 1/16” legs, 160 X strips=320 joints = 3.2gr of glue; so bracing wood + glue = 7.12gr
For the 3/32” legs, 72 X strips=144 joints = 1.44gr of glue; so bracing wood + glue = 3.21gr
For the 1/8” legs, 40 X strips=80 joints = 0.8gr of glue; so bracing wood + glue = 1.79gr
And finally, adding it all together,
For the 1/16 legs 7.12gr bracing+0.35gr in legs = 7.47gr segment weight
For the 3/32 legs 3.21gr bracing+0.71gr in legs = 3.92gr segment weight
For the 1/8 legs 1.42gr in bracing+1.79gr in legs = 3.21gr segment weight.
Finally, a pretty clear answer to which is better.