Tetra Math

[ZZII 527]

Not that you all couldn't figure this out on your own, but I'll spare you the effort:

For a tetrahedron with side of length x:

distance from midpoint of one side to opposite corner of same face (base altitude): sqrt(3)*x/2

distance from midpoint of one side to the horizontal center of the tetra (below the top point): sqrt(3)*x/6 (1/3 total length of base altitude)

distance from any base point to the horizontal center of the tetra (below the top point): sqrt(3)*x/3 (2/3 total length of base altitude)

height of tetra: sqrt(2/3)*x

[Levin571]

does anyone know how many tetra's can be fit under one of the goal by any chance?

[tkwetzel]

Originally Posted by Levin571

does anyone know how many tetra's can be fit under one of the goal by any chance?

A lot of it depends on how they are stacked and whatnot. Because they could be stacked like they are on the top, but they can also just be pushed in next to each other. One person will post one number only to have 4 other people post 4 different numbers. Though I think your best bet on getting more of them under there would be stacking them. Also, without stacking and just looking at the dimensions, you should be able to fit four COMPLETELY inside and flat on the ground within a singel goal. Please note that the scoring objects do not have to be completely within the goal to count though. It will count as long as it breaks the plane of the goal and it is not supported by the carpet outside the goal or a robot.

[Eugenia Gabrielov]

Originally Posted by Levin571

does anyone know how many tetra's can be fit under one of the goal by any chance?

However, keep in mind that in order to stack tetra's under your goal, it takes more time to obtain them and manuever your arm to make it perfect then it does with stacking them on the goal (same amount of effort) and also you get more points for stacking and owning. Containing gives you 1 points each, if I am correct in remember stacking gives you 3. Before desigining your robot, remember that some point values are crucial in a game with such close balances of points.

[n0cturnalxb]

Originally Posted by Levin571

does anyone know how many tetra's can be fit under one of the goal by any chance?

Considering the goals (aside from the center one) are about 5'3" = 63"(.. that's almost as tall as I am! .. ugh, I feel short now) and each tetra is 2'4" = 28" ...

Let's say each tetra raises the stack about 3"?

And you need about 28" clearance left to fit the last tetra in ...

4? That doesn't sound right. 7ish-9ish?

I think I just fried my sleep-deprived brain.


(edit) Err, cross that out .. I mean around 4-6 maybe per stack.. maybe.. assuming each tetra raises each stack 3". I think. Someone check my math or logic or something, brain's not working well right now..

Assuming you can fit 4 perfect stacks of 4 inside the goal, that's 16.

I'd say go with stacking on top of the goal.. go for containing if you're having a lot of problems with designing the arm, but stacking will give you three times the amount of points you can get by containing .. and the possibility of making a row.

[Dillon Compton]

Originally Posted by n0cturnalxb

Considering the goals (aside from the center one) are about 5'3" = 63"(.. that's almost as tall as I am! .. ugh, I feel short now) and each tetra is 2'4" = 28" ...

Let's say each tetra raises the stack about 3"?

And you need about 28" clearance left to fit the last tetra in ...

4? That doesn't sound right. 7ish-9ish?

I think I just fried my sleep-deprived brain.


(edit) Err, cross that out .. I mean around 4-6 maybe per stack.. maybe.. assuming each tetra raises each stack 3". I think. Someone check my math or logic or something, brain's not working well right now..

Assuming you can fit 4 perfect stacks of 4 inside the goal, that's 16.

I'd say go with stacking on top of the goal.. go for containing if you're having a lot of problems with designing the arm, but stacking will give you three times the amount of points you can get by containing .. and the possibility of making a row.

According to the kickoff, each tetra raises the stack approx. six inches.

[sojouner06]

one tetra added to the goal is a height increase of exactly 3.5 inches. Therefore, if you just add about 5 tetra's, thats about another foot and a half. So if you have an arm, you would need to have a 10.5 foot at full extension to out class other arms

[Max Lobovsky]

Originally Posted by sojouner06

one tetra added to the goal is a height increase of exactly 3.5 inches.

Where did you get that? From pictures (and according to Dillon Compton) it looks much closer to 6" per tetra.

[Ryan F.]

It is 6 inches per tetra...this is comming from someone who's seen them built and stacked on eachother.

[sojouner06]

i calulated using a formula i learned in statistics, the width of the pole vs height/ complicated math suff

[sojouner06]

pole=pvc pipe

[sojouner06]

i made tetra's too and did see that it was 3.5, ill go see if the tetra's was built properly

[elknise]

sojouner06, try to edit your posts instead of creating new ones.
ftp://67.170.35.253/robotics/Posting.swf

Anyways, did you take into consideration about the end-caps. Also, when looking at the sidelines during the kickoff, it looked more like 6 inches.

[HPA_Robotics_13]

I thought that the tetras used at the actual competitions don't have PVC endcaps but instead have flat pieces that screw into the PVC?

[Greg Needel]

i was in manchester and saw the offical ones and adding them ontop of each other adds SIX inches....i also tested the how many fit under a goal and 7 perfectly stacked will make it as long as there are no tetras ontop of the goal. if there is a tetra ontop of the goal then you can only fit 4 under