Tribal Territory Size Answer

[Oakstead]

Both Farmerbob's formula and pid73's formulas are correct (you beat me too it!).

This produces the following results:

10 player tribe = 7,850 square meter area (100 meter diameter)
20 player tribe = 17,662 square meter area (150 meter diameter)
30 player tribe = 31,400 square meter area (200 meter diameter)
40 player tribe = 49,062 square meter area (250 meter diameter)
50 player tribe = 70,650 square meter area (300 meter diameter)

This scheme for tribal areas encourages tribes to have only one identity instead of breaking up into sub-tribes with each having their own area because a 20 player tribe will have more area then two 10 player tribes. This pattern continues no matter the number of players in a tribe.

Edited to say pid73's formula is correct and just a different version of Farmerbob's

[Jadzia]

I wonder if there is a limit for the population of tribes.

[pid73]

Alfred wrote:

T = pi * (50 + 5 * n)^2 / 4

It's actually: pi * (50 + 2,5 * n)^2 because an addition to the diameter is always half the addition to the radius. Since the equation for the area of a circle is pi*r^2, we are using the radius and the division by 4 is a mistake.

It still means that to dominate the world (120km^2) you will need about 760 people in the same tribe.

Are you really sure about what you said?
Why don't you open up Excel and run some numbers?

for example:
20 members raise the diameter to 150m (100m + 50m). That's a radius of 75m, encircling an area of 17671 m^2.

That's exactly what I get with my formula.

Your formula for 20 members is wrong. I punch the number 20 into your formula and get:

pi * (50 + 2,5 * 20)^2

That is a radius of 100m, which means the diameter should be 200m. But 20 members do not give you a diameter of 200m, but 150m.

[pid73]

Oakstead wrote:

Farmerbob's formula is correct (you beat me too it!).

This produces the following results:

10 player tribe = 7,850 square meter area (100 meter diameter)
20 player tribe = 17,662 square meter area (150 meter diameter)
30 player tribe = 31,400 square meter area (200 meter diameter)
40 player tribe = 49,062 square meter area (250 meter diameter)
50 player tribe = 70,650 square meter area (300 meter diameter)

This scheme for tribal areas encourages tribes to have only one identity instead of breaking up into sub-tribes with each having their own area because a 20 player tribe will have more area then two 10 player tribes. This pattern continues no matter the number of players in a tribe.

Do you see any difference with my forumla?
I don't. The numbers are the same.

I just take steps of 5 members while Farmerbob takes steps of 10 members.

[Derek]

Guys, you misunderstood me. Btw, your both talking about the same formula A=pi*D^2/4=pi*r^2.

What I meant was that each additional tribe member should add less and less to the diameter, diminishing returns.

[Oakstead]

You're right Pid73, your formula is correct as well. Sorry for not seeing that in my rush to get the data up.

[Farmerbob]

Derek wrote:

Guys, you misunderstood me. Btw, your both talking about the same formula A=pi*D^2/4=pi*r^2.

What I meant was that each additional tribe member should add less and less to the diameter, diminishing returns.

Your heart is in the right place but:

Assume R = D/2

D^2/4 is not equivalent to R^2 over any range.

I think you meant A = pi*((D/2)^2) = pi*(R^2)

I did not look back to verify, I'm only going by what I see above.

[Farmerbob]

Jadzia wrote:

I wonder if there is a limit for the population of tribes.

I strongly doubt that any tribe large enough to actually control a significant area on the server will be able to maintain cohesiveness. The biggest corporation in EVE that I can recall was EVE University, a training corp, and I do not believe their membership ever exceeded 2000 players, with an active playerbase of 200,000+ Alliances on the other hand got rather larger at times, but they are not single political units.

Sure Xsyon is not EVE, but they are both Sandbox, and share a lot of game dynamics, even if the background for said dynamics is drastically different.

[pid73]

Oakstead wrote:

You're right Pid73, your formula is correct as well. Sorry for not seeing that in my rush to get the data up.

Don't be sorry, maybe I sounded wrong in the previous post.
I wanted to point it out but was by no means "offended"...
I had to put this in that post: :silly:

[Derek]

Farmerbob wrote:

Derek wrote:

Guys, you misunderstood me. Btw, your both talking about the same formula A=pi*D^2/4=pi*r^2.

What I meant was that each additional tribe member should add less and less to the diameter, diminishing returns.

Your heart is in the right place but:

Assume R = D/2

D^2/4 is not equivalent to R^2 over any range.

I think you meant A = pi*((D/2)^2) = pi*(R^2)

I did not look back to verify, I'm only going by what I see above.

Ok, I don't mean to get into a pissing contest over something so trivial, but I use that formula every day at work (I'm a structural engineer). Your not reading it correctly A=pi*(D^2)/4=pi*r^2