**First, forces on a leg.**For discussion, lets assume a 4-leg tower. Having not gotten this year’s rules yet, don’t know what the size of the hole in the test base the legs need to clear/be outside of.

So, let’s start with a simple case- a … hypothetical straight tower; legs vertical, on the corner of a 5cm square (so the 5cm square load block just sits/fits on top). With a 15kg load, each leg will carry 1/4th of that load- a 3.75 kg “axial compressive” force (i.e., the load puts a compressive force along the long axis of each leg). If the force is strong enough, or the leg weak enough, it will fail by “buckling”- the middle will bow outward due to the force, and the leg will break.

So what happens/what’s the math, if the legs are angled (wider at the base/bottom- to clear the defined base opening)? As discussed above, the “more angled” the legs are, the greater the force they have to carry (at a given load). If you do a vector analysis, you’ll find that force has an “inverse cosine “ relationship to the angle from vertical- the force is 1 over the cosine (1/cos) of the angle from vertical, times the load. If you drop an imaginary vertical line from the top of one leg down to the test base, you form a right triangle. The hypotenuse of that triangle is the leg. The imaginary vertical line is the other (adjacent) side of the angle from vertical. The cosine of the angle between the hypotenuse and the adjacent side is the length of the adjacent side divided by the length of the hypotenuse.

Looking at a few cosine values, 1/cos, and axial compressive force on one leg of a four leg tower with a 15 kg load

0 degrees; cosine = 1.0000; 1/cos = 1.000; axial compressive force = 3.75 kg

5 degrees; cosine = 0.9962; 1/cos = 1.0038; axial compressive force = 3.78 kg

10 degrees; cosine = 0.9847; 1/cos = 1.0155; axial compressive force = 3.81 kg

15 degrees; cosine = 0.9661; 1/cos = 1.0351; axial compressive force = 3.88 kg

20 degrees; cosine = 0.9397; 1/cos = 1.0642; axial compressive force = 3.99 kg

30 degrees; cosine = 0.8658; 1/cos = 1.1550; axial compressive force = 4.33 kg

The bottom line, until that angle gets pretty big, the effect of the angle on the force on the leg is pretty small. Even at 30 degrees from vertical, the increase in force (from that on a vertical leg) is about 15%

**The second important relationship to understand is how the strength of a leg (resisting buckling) relates to its length.**This is an “inverse square” relationship. Google up Euler’s Buckling Theorem to see the formula. What its saying is that if you have a column- a leg- a piece of wood say 60cm long, and it has a buckling strength of 0.5 kg, if you cut its length to 1/2, to 30 cm, its buckling strength goes up by a factor of 4 (to 2.0 kg); if you cut its length to 1/3, to 20 cm, its buckling strength goes up by a factor of 9 (to 4.5 kg).

You can see/get a feel for this important relationship by taking a (nice straight) balsa stick, place it on a scale, vertically, and push straight down. At some force/load, the middle will start to bow out (buckle)- that’s its buckling strength at that length. Be careful not to break it; just note/see the force/weight measured by the scale when it starts to buckle. Now, cut the stick in half, and do the same “axial loading” to the point buckling starts, and you’ll see the force on the scale is about 4 times what it was at full length. The numbers you get won’t be perfect; there will be some variation in the stiffness along/within the stick (that’s just the nature of wood- its not perfect/homogeneous), and its likely not perfectly straight, and without a pretty precise testing setup, the force you put on the top won’t be perfectly vertical, but you will see pretty close to 4 times the buckling strength.

What the bracing between the legs is all about is turning the long leg/column (with a low buckling strength) into a set of shorter, equal length, “stacked columns” (with much higher buckling strength). You can get a feel for this by repeating the ‘push down on a long stick on the scale” drill. Mark the midpoint on the stick- push down at full length, note the weight/load at which buckling starts. Now, push it to the start of buckling again, but with your other hand, hold the midpoint in place, so it doesn’t move in space, and what you’ll see is that at about 4 times the long length buckling load, one of the sections (above or below your fingers holding/bracing the midpoint) will start to buckle. There is good, detailed discussion in the forum archives on approaches to doing the bracing. Key is keeping the “braced interval” equal- if the exposed section of the leg between braced points is unequal, the longer exposed leg segment will break first.

**The third/last important relationship to understand is the effect of density on the stiffness of a leg.**This is pretty close to a linear relationship; if you have two sticks w/ the same size/cross section (e.g. 1/8” x 1/8”), and one weighs twice what the other weighs, the heavier one will have about twice the buckling strength at a given length. ( http://ir.library.oregonstate.edu/xmlui ... /1957/1286 )

Understanding these basic relationships gives you the basis for designing an efficient tower.