Towers B/C

[Balsa Man]

I know its been a while. For the solvent you (Balsa Man) showed me.

https://www.amazon.com/dp/B00JFPF0UQ/re ... ullets-btf.

Does it also bond with the edge strips. I'm thinking of using PLA or ABS plastic for the edge strips.

Yes

[Balsa Man]

Ok, I though the answer would be more simple than that. It was more of a general question. I don't have any specific weights or cross sections to give you.

There is a VERY simple way to start to understand the answer to your general question; test. Take 2 sticks that meet the conditions you stated, and see which has higher BS.

The issue your question goes to is interesting, and worth understanding. I'll get a post together that discusses how you can figure out how cross section and density affect BS. In general, as you increase cross section, buckling strength goes up at a faster rate than stick weight does.

[Girlpower05]

What bracing intervals would work best for a division B tower this year?

Yes, you have summed up the entire tower competition in one sentence!
The bracing interval is a function of the column density and stiffness. As the column density goes up, the number of bracing tiers can come down (theoretically, anyhow). The lighter the columns, the more bracing will be required.

First, just for clarity, when you say “work best”, I’m assuming you mean will get you the highest scoring tower. And yup, figuring out what will work best is the problem; the challenge; what this game is all about.

Second, dholdgreve is describing the fundamental tradeoff you have to understand and figure out to answer the question of what will work best. It is a tradeoff, because with stiffer, heavier, stronger leg wood, the bracing interval that will work (where ‘work’ means the leg segment has been braced enough to carry your design load) can be longer, more open, meaning you’ll have less wood weight in your braces. Conversely, with floppier, lighter, weaker leg wood, the bracing interval that will work has to be tighter, shorter, meaning you’ll have more wood weight in your bracing. The type of bracing also affects the bracing interval that will work with legs of various strength.

I’m not going to just give you the answer (or what I think the answer is) to your question, but I’m more than glad to share some basic information, and a couple of engineering tools that will give you what you need to work out the answer your question. It’s like the difference between giving someone a fish, and teaching them how to catch a fish. What you need to be able to figure out, and look at, and compare is the weight of the leg wood + the weight of the bracing wood + the weight of glue needed, for a range of bracing intervals. With this year’s ‘2-part’ rules, you actually need to look at this tradeoff separately for the upper/chimney section, and the lower/base section.

For each section, this is a 2-part problem. The first problem is a strength problem; the second problem is a density problem.

Looking at the strength problem first, when I say strength, I’m talking about buckling strength. “Stiffness” is… another way of saying this. How much force does it take for a braced segment of a leg to start to bow out in the middle (which is what buckling is)? - stiffer means has higher buckling strength. Buckling is how axially loaded thin columns fail; the legs in a tower are axially loaded thin columns. Once buckling/bowing starts, its all over. With the same force that started the bowing of a leg segment being applied, the bowing will progress… very rapidly, and the segment will snap.
We know buckling strength has an “inverse square relationship” to length. What is that saying? When you brace a column (a leg section), you create a set of shorter ‘stacked columns.’ If a full leg segment (be it chimney or base segment) at length L has a buckling of X, if you brace it at the middle, you have two shorter columns/braced intervals with a length of L/2 (the proportion of the braced intervals is ½), and a buckling strength (BS) of 4X. If you brace it into thirds (a 1/3 bracing interval – a proportion of 1/3), the three braced intervals will have a buckling strength of 9X, and so on. The general form we see here is that the BS of a shorter braced interval is 1 over, as in 1 divided by, the proportion squared; that an inverse square relationship. “Proportion” is the proportion of the shorter interval to the longer interval.

As discussed many times, we can measure the BS of a stick (like a 36” stick), and do an inverse square calculation to know what the BS will be increased to at a given bracing interval. One could do a bunch of inverse square calculations, looking at a range of stick buckling strengths, and bracing intervals, to see what bracing interval would be needed, for what stick buckling strength, to get a leg strength that would work- as in carry the force the leg will see when the tower is loaded. A spreadsheet with a range of braced intervals, and a range of measured stick BSs, calculating the braced BS for a given bracing interval, and given stick BS is the best tool for doing this. With it, you can solve the strength problem- to know what strength wood you would need, to have various bracing intervals “work.”

On to the density problem; with a way to know what strength leg wood you will need to use each of a range of bracing intervals, we now need a way to figure out what weight/density of wood it will take to have, to get, a given BS. If one had data on a whole bunch of balsa sticks – for each stick, the stick weight, and the measured BS, and you put that into a graph plotting stick weight vs BS…..what could one do? You could look in your inverse square table and, for each of the bracing intervals you’re evaluating, see what stick BS you need, and look at the graph of stick weight vs BS and see the stick weight you would need to have the BS you need. This sort of process, and development and use of tools is at the heart of ‘”engineering,” and “the design process.”

These two basic tools you will give you what need to evaluate the tradeoff we started this discussion with. With stick weight, you’ll be able to calculate the leg weight (that would work) for each bracing interval. And, you can easily figure out the length of the bracing pieces needed for each bracing interval (by just measuring them off a scale drawing)- then you’ll know how many total cm of bracing wood needed for each bracing interval being evaluated.

Then, with bracing wood length, you can calculate bracing wood weight (by using a density value and cross sectional size for the bracing). Actually, you’d want to use two or three values; a light, medium, and high, and calculate the grams per cm of bracing wood for each of the three scenarios. Gr/cm times total bracing length (for each of the bracing interval scenarios) gives you weight of bracing wood. If you’re using an “all Xs” bracing approach, with 1/32” x 1/16” (‘the hot setup’ from last year. And no reason to believe it won’t be this year), light would be about 6gr 1/32” x 3” x 36” sheet, medium about 7.5gr sheet, and heavy about 9gr). Last factor is glue weight; you need a little bit of glue at each end of each bracing piece. A workable number for the evaluation we’re trying to do is 0.03gr/joint.

So, what are you going to have to do to get the two basic tools I described?.

Go back to page 6 of this thread. See the links in my long my post on Oct 20, near the top of page 6. Read the post, click on the links, study, and enjoy.

The inverse square table is set up for all Xs using 1/32 x 1/16 Xs. If you want to look at ladders and Xs, you need to change the effective length factor in cell N8 to 2.3 (and add lines to look at wider bracing intervals). The post discusses what the effective length factor is all about.

Thank you for the detailed response! I read through the other post and understand what to do.

[Balsa Man]

Thank you for the detailed response! I read through the other post and understand what to do.

Excellent; you're on your way! Glad this was helpful, and hope you can do well in this event. Go for it!

[Random Human]

Just a note to everyone. If you have a question, go to the 2017 thread. The threads are practically a field guide towards the victory. I guarantee that 95% of the questions you have, will be answered in the 2017 thread.

[AAH2020]

Do any of you have ideas for a jig for this year's design? The biggest problem I'm stumped at is designing a jig that allows accurate attachement of the diagonal cross pieces that connect the base to the main tower. I'm just very stumped on how to cut blocks of wood that allow us to place our main tower section on one piece and connect it with precision to our 30cm base square.

[Balsa Man]

Do any of you have ideas for a jig for this year's design? The biggest problem I'm stumped at is designing a jig that allows accurate attachement of the diagonal cross pieces that connect the base to the main tower. I'm just very stumped on how to cut blocks of wood that allow us to place our main tower section on one piece and connect it with precision to our 30cm base square.

Not trying to be rude, but if you had read what is already here, in this thread, you would find exactly what you're asking about, discussed in some detail (in this thread, and the new thread you created to ask the same question), and you would know that for a base meeting the 29cm circle bonus- 4-leg, square base tower, you're looking at just a bit over 21cm square, not a 30 cm square. Draw a 29cm circle. draw a square that has its corners just outside the circle. Measure the sides of that square.
Respectfully,

[AAH2020]

That's not what I meant at all. With all due respect, I already knew the exact specifications and such. However, we are wanting to use a more unique design that uses just one main beam for each corner, as opposed to 3-4 side pieces combined together for maximum weight conservation. I was wondering if anyone has made a regular wood jig that allows an accurate glueing for the base and thin sections for an accurate angle joint. I
I will try to post a picture of our goal design in a couple of days, noting where we're having trouble with.
Regardless, thank you for noting my question and responding.

[Random Human]

That's not what I meant at all. With all due respect, I already knew the exact specifications and such. However, we are wanting to use a more unique design that uses just one main beam for each corner, as opposed to 3-4 side pieces combined together for maximum weight conservation. I was wondering if anyone has made a regular wood jig that allows an accurate glueing for the base and thin sections for an accurate angle joint. I
I will try to post a picture of our goal design in a couple of days, noting where we're having trouble with.
Regardless, thank you for noting my question and responding.

I'm not really sure of the motives for saving weight on the jig?? I'm kind of confused by your question...

[BananaPirate]

Then, with bracing wood length, you can calculate bracing wood weight (by using a density value and cross sectional size for the bracing). Actually, you’d want to use two or three values; a light, medium, and high, and calculate the grams per cm of bracing wood for each of the three scenarios. Gr/cm times total bracing length (for each of the bracing interval scenarios) gives you weight of bracing wood. If you’re using an “all Xs” bracing approach, with 1/32” x 1/16” (‘the hot setup’ from last year. And no reason to believe it won’t be this year), light would be about 6gr 1/32” x 3” x 36” sheet, medium about 7.5gr sheet, and heavy about 9gr). Last factor is glue weight; you need a little bit of glue at each end of each bracing piece. A workable number for the evaluation we’re trying to do is 0.03gr/joint.

Hi Balsa Man,
Here you mentioned that we would want to consider 2-3 types of sheets (difference in weight) for the X bracing, which makes sense. However, I'm confused on how the mass distinctions you made (6, 7.5, 9g) are actually used to make a decision without actually testing all 3 types, possibly more for sheets in between these 3, as I do not see a way to mathematically figure out if each density bracing will be able to hold the compression and tension forces. I assume that I'm missing a key assumption? Thanks for the help!