Tetra Math [Archive] - Chief Delphi

ZZII 527

09-01-2005, 00:31

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

09-01-2005, 00:33

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

tkwetzel

09-01-2005, 00:40

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

09-01-2005, 00:42

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

09-01-2005, 00:43

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

09-01-2005, 12:21

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

10-01-2005, 00:08

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

10-01-2005, 00:11

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.

10-01-2005, 00:12

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

sojouner06

10-01-2005, 00:14

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

sojouner06

10-01-2005, 00:15

pole=pvc pipe

sojouner06

10-01-2005, 00:19

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

elknise

10-01-2005, 01:54

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

10-01-2005, 01:58

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

10-01-2005, 02:05

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

tkwetzel

10-01-2005, 11:13

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?

The small tetras have flat clover pieces that screw into the end caps attached to the PVC. The goals are aluminum and will have a angled piece of hardware holding them together. These pieces are visible in the field drawings available on FIRST's documents page.

Aignam

10-01-2005, 11:14

The small tetras have flat clover pieces that screw into the end caps attached to the PVC. The goals are aluminum and will have a angled piece of hardware holding them together. These pieces are visible in the field drawings available on FIRST's documents page. In addition, a good picture of these pieces can be located here (http://www.chiefdelphi.com/forums/pictures.php?action=single&picid=9188).

Lucid

10-01-2005, 11:17

anyone know what the angle is between faces? I drew a tet in AutoCad and I measured it to be 55 degrees, but I think there is something wrong with this.

Lucid

Grommit

11-01-2005, 02:18

The angle can be calculated by looking at a cross section of the tetra, giving 2*arcsin(1/sqrt(3)), which is approximately 70 degrees.

Leav

11-01-2005, 05:09

The angle can be calculated by looking at a cross section of the tetra, giving 2*arcsin(1/sqrt(3)), which is approximately 70 degrees.

I got a much simpler expression: cos^-1(1/3) whuch equals 70.528779365509308630754000660038

:)

I love math, algebra and trig :D

me = nerd :p

unapiedra

11-01-2005, 18:25

if there is a tetra ontop of the goal then you can only fit 4 under

Why that? Does the tetra ontop block that?

sojouner06

11-01-2005, 22:40

think about it here, 6 inches, thats half a foot. Even with end caps at the largest constraint is 4 inches. It is physicaly impossible for it to be a six inch increase per tetra. IMPoSIBLE!!!!!!!!!!!! unless someone put it on a freaky way

ZZII 527

12-01-2005, 05:29

We've also built tetras...it is 6". The sides of one tetra do not even come close to being flush up against the sides of the one below it. I think it is partially because the sides of the lower one get jammed into the corners of the one on top.

Twstdkittie

12-01-2005, 09:20

rangersteve

12-01-2005, 18:34

I wish i had seen this thread sooner. I had to do this math today but oh well i like math. Nice thread.

sojouner06

12-01-2005, 22:26

ok i believe it is 6 inches, but wait, how thick are your pipes. The manual said they should be 1 and a quarter inches at pipe and 1.(4-9) inches at endcap

or...maybe...how long are the pipes themselves? 40"? 48"? 70"?

sojouner06

12-01-2005, 22:29

http://www.ifirobotics.com/first-2005-field-parts.shtml

look here for the parts first wants you to get

tkwetzel

12-01-2005, 22:53

http://www.ifirobotics.com/first-2005-field-parts.shtml

look here for the parts first wants you to get

With the correct parts, the tetras will add approx. 6" each to the height of the goal when stacked.

sojouner06

13-01-2005, 18:17

ok, it is 6 inches, but i am still baffaled! how can the formula i used be wrong???? what could be an outside influence to change the height so drastically...

ZZII 527

13-01-2005, 22:04

To clarify angles a bit:

The angle between a face and an edge is ~55 degrees. The angle between two faces is ~70 degrees.

Twstdkittie

14-01-2005, 08:47

ok, it is 6 inches, but i am still baffaled! how can the formula i used be wrong???? what could be an outside influence to change the height so drastically...

Because the angle at the top of the small tetras isn't big enough to completely go over the angle on the big one.

Dr.Bot

14-01-2005, 09:07

The tetrahedral arangement of carbon atoms is what gives diamond the properties of density, hardness, and optical index of refraction(sparkle).

A tetrahedron can be made from an equilateral triangle (All angles 60 degrees)
If you start with a triangle and then bisect eact side, You will have 4 triangles. By folding each face up, you will have a tetrahedron. You can see how that even as a solid shape these can pack together perfectly. In crystals this usually means high density, and isotropic ( the same in all directions) properties.

Pingponger17

18-01-2005, 09:18

I haven't seen two actually stacked so I don't know for sure how much of a raise each tetra adds. But in the rules i believe it said that if a tetra is stacked improperly and has more than 6 inches then it isn't considered to be stacked. So i believe that it raises less than 6 inches to leave a little margin for error.