I've only got the demo, and only done each level once, so I dont have so many balls,but I got 18.9 meters tall, using 84 balls, that any good?
Really enjoying the game, going to get full version on pay day ^^ 
:o
I can barely reach that height with 240 balls !!!
If you can continue with the full version of the game, you may well take the "Championship of the World" in gooball tower building. 
That's pretty good. :)
I think that as a rule of thumb, your tower is impressive if the height you reach is more than twice the square root of the number of balls you use.
So players using 300 balls (the max) have an impressive tower if it is over 34.6m tall.
With 84 balls, a tower over 18.3m is impressive. 
But.... how the hell can I ever reach 3.5 meters with my 3 balls? ::) 
Hmm, interesting challenge, but given that you've managed something like 12m with 23 balls, I'm sure you'll manage ;D
But anyway, a tower over 3.5m with 3 ball would definitely be "impressive", wouldn't it? :)

On a more serious note:
To build a basic equilateral triangle with sides of length X (X edges between X+1 nodes) you need X^2 + 2 goo. So the triangle you start with at WoG Corp has side 1 and takes 3 goo. Making it an equilateral triangle with sides of length 2 takes 3 additional goo for a total of 6, etc.
This means that with X balls, you can easily build a triangle with sides of length sqrt(X2). Such a a triangle will be roughly X meters high (a little higher, actually, something like 1.15 * X). So a height of sqrt(X2) is "easy". (So with 84 balls, 9m is "easy").
For small numbers, the 2 in there becomes important. So actually, it would be more accurate to use 2 * SQRT(X2) as the heigth for impressive towers.
To improve on the equilateral triangle, you need a structure with steeper sides. My estimate of "twice the square root" was based on the current leaderboard, which has only 8 towers that exceed twice the square root of 300.
Last modified Tue, 11/04/2008  07:14 by herminator 
[quote author=herminator link=topic=700.msg5104#msg5104 date=1225800648] To build a basic equilateral triangle with sides of length X (X edges between X+1 nodes) you need X^2 + 2 goo. So the triangle you start with at WoG Corp has side 1 and takes 3 goo. Making it an equilateral triangle with sides of length 2 takes 3 additional goo for a total of 6, etc. It seems you are only counting the number of nodes in the triangle; since the triangle has more than twice the number of edges, you need to use goos for edges too. You do seem to arrive at the right conclusion though.
A tower of height N (where N is the number of levels; so a simple triangle with what you call side 1, has height 2) has N*(N+1)/2 nodes and 3*N*(N1)/2 edges. If start with three edges and three nodes (the initial triangle) you can subtract 3 from both of these. This means that for a tower of height N, you need to add 3*N*(N1)/2  N*(N+1)/2 = N*(N2) goos (or N^{2}  2N if you prefer that notation).
In other words, with N balls you can build a triangle of height floor(1 + sqrt(1 + X)) (sqrt being the square root; floor rounding down to the nearest integer); in your terminology that would be a triangle with sides length floor(sqrt(1 + X)).
Quote: This means that with X balls, you can easily build a triangle with sides of length sqrt(X2). Such a a triangle will be roughly X meters high (a little higher, actually, something like 1.15 * X). I assume you don't mean "X meters high" but H, where H ~ sqrt(X). Based on a nominal edge length of 140 mm, each row in the triangle is 1.40*sin(60) = 1.40*sqrt(3/4) ~ 1.2124. So with X balls, the easy height should be around 1.4*sqrt(3/4)*floor(sqrt(1 + X)), not accounting for edges on the bottom being compressed.

[quote author=Soultaker link=topic=700.msg5122#msg5122 date=1225817402] [quote author=herminator link=topic=700.msg5104#msg5104 date=1225800648] To build a basic equilateral triangle with sides of length X (X edges between X+1 nodes) you need X^2 + 2 goo. So the triangle you start with at WoG Corp has side 1 and takes 3 goo. Making it an equilateral triangle with sides of length 2 takes 3 additional goo for a total of 6, etc. It seems you are only counting the number of nodes in the triangle; since the triangle has more than twice the number of edges, you need to use goos for edges too. You do seem to arrive at the right conclusion though.
Erm, no, I was counting the edges too, which is why I arrived at the right conclusion. The number of nodes would be ((X+1)(X+2))/2.
Quote: A tower of height N (where N is the number of levels; so a simple triangle with what you call side 1, has height 2) has N*(N+1)/2 nodes and 3*N*(N1)/2 edges. If start with three edges and three nodes (the initial triangle) you can subtract 3 from both of these. This means that for a tower of height N, you need to add 3*N*(N1)/2  N*(N+1)/2 = N*(N2) goos (or N^{2}  2N if you prefer that notation).
In other words, with N balls you can build a triangle of height floor(1 + sqrt(1 + X)) (sqrt being the square root; floor rounding down to the nearest integer); in your terminology that would be a triangle with sides length floor(sqrt(1 + X)).
Yes, you can count the length of the side of a triangle in either nodes or edges, but the calculations basically stay the same.
Quote:
Quote: This means that with X balls, you can easily build a triangle with sides of length sqrt(X2). Such a a triangle will be roughly X meters high (a little higher, actually, something like 1.15 * X). I assume you don't mean "X meters high" but H, where H ~ sqrt(X). Based on a nominal edge length of 140 mm, each row in the triangle is 1.40*sin(60) = 1.40*sqrt(3/4) ~ 1.2124. So with X balls, the easy height should be around 1.4*sqrt(3/4)*floor(sqrt(1 + X)), not accounting for edges on the bottom being compressed.
Yes, sorry, height H ~ sqrt(X). 