What's the highest value for "influence park"?

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starguything
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What's the highest value for "influence park"?

#1

Post by starguything »

I tried making a 6x6 park with an "influence park" value of somewhere along the lines of 2043 which came from a very rough calculation. I tried it out and noticed that the influence of my park and the Big park in-game, which I think has the highest influence out of all the other default parks, have the same influence: 16 tiles next to each edge of the park at their greatest extent. I wanted to make a park with an influence that spans 51 tiles in diameter but as from what you can see it just ends up being 38 tiles in diameter.

EDIT: Whoops. Apparently the JSON code I copied and pasted from and built upon had "inlfluence park" on it instead of "influence park". I also observed getting the influence you want didn't even really need calculation; the influence value is pretty much the number of tiles the building can influence from its center point. In my case, I input the value 29; 3 for the tiles in the building from its center and 26 beyond. I appreciate the help :).
Last edited by starguything on 05 Aug 2018, 22:58, edited 1 time in total.

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Re: What's the highest value for "influence park"?

#2

Post by CommanderABab »

The highest in game park is 30 except for the mountain. You might try 100.

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Re: What's the highest value for "influence park"?

#3

Post by Lobby »

A tile that provides an influence value of \(i\) has the influence effect \(f\) on another tile at relative position \(\Delta x, \Delta y\):
\(f(\Delta x,\Delta y,i)=\frac{i}{\Delta x^2+\Delta y^2+1}\)
with 1.0 being the max and 0.0 being the min value for influence effect.

So if you want a single tile to have a specific effect \(e\) at a distance \(d=\sqrt{\Delta x^2+\Delta y^2}\) you can calculate the required influence value \(i\) by
\(e=\frac{i}{\Delta x^2+\Delta y^2+1}\\
\Leftrightarrow i=e\cdot(\Delta x^2+\Delta y^2 + 1)\\
\Leftrightarrow i=e\cdot(d^2+1)\\
\Leftrightarrow i =e\cdot d^2+e\)


So your rough calculation is quite good, at least for a single tiles. To get an influence effect of \(0.5\) at a distance of 50 tiles you would need an influence value of
\(i=0.5\cdot2,500+0.5\approx1,250\).
But all of your tiles (36 for a 6x6 sized building) have an influence effect, so in reality a value around\(\frac{2,500}{36}\approx70\) seems to be reasonable (neglecting the different distances of the tiles of the building).

The precision for influences is limited to reduce memory consumption. Therefore huge numbers (like 5000 :lol:) may lead to an overflow which in return may lead to strange influence effects.

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Re: What's the highest value for "influence park"?

#4

Post by CommanderABab »

Lobby wrote:
05 Aug 2018, 09:15
A tile that provides an influence value of \(i\) has the influence effect \(f\) on another tile at relative position \(\Delta x, \Delta y\):
\(f(\Delta x,\Delta y,i)=\frac{i}{\Delta x^2+\Delta y^2+1}\)
with 1.0 being the max and 0.0 being the min value for influence effect.

So if you want a single tile to have a specific effect \(e\) at a distance \(d=\sqrt{\Delta x^2+\Delta y^2}\) you can calculate the required influence value \(i\) by
\(e=\frac{i}{\Delta x^2+\Delta y^2+1}\\
\Leftrightarrow i=e\cdot(\Delta x^2+\Delta y^2 + 1)\\
\Leftrightarrow i=e\cdot(d^2+1)\\
\Leftrightarrow i =e\cdot d^2+e\)


So your rough calculation is quite good, at least for a single tiles. To get an influence effect of \(0.5\) at a distance of 50 tiles you would need an influence value of
\(i=0.5\cdot2,500+0.5\approx1,250\).
But all of your tiles (36 for a 6x6 sized building) have an influence effect, so in reality a value around\(\frac{2,500}{36}\approx70\) seems to be reasonable (neglecting the different distances of the tiles of the building).

The precision for influences is limited to reduce memory consumption. Therefore huge numbers (like 5000 :lol:) lead to an overflow which in return may lead to strange influence effects.
Where may one find latex language documentation?

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Re: What's the highest value for "influence park"?

#5

Post by Lobby »


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Re: What's the highest value for "influence park"?

#6

Post by LobbyFan »

Lobby wrote:
05 Aug 2018, 09:15
A tile that provides an influence value of \(i\) has the influence effect \(f\) on another tile at relative position \(\Delta x, \Delta y\):
\(f(\Delta x,\Delta y,i)=\frac{i}{\Delta x^2+\Delta y^2+1}\)
with 1.0 being the max and 0.0 being the min value for influence effect.

So if you want a single tile to have a specific effect \(e\) at a distance \(d=\sqrt{\Delta x^2+\Delta y^2}\) you can calculate the required influence value \(i\) by
\(e=\frac{i}{\Delta x^2+\Delta y^2+1}\\
\Leftrightarrow i=e\cdot(\Delta x^2+\Delta y^2 + 1)\\
\Leftrightarrow i=e\cdot(d^2+1)\\
\Leftrightarrow i =e\cdot d^2+e\)


So your rough calculation is quite good, at least for a single tiles. To get an influence effect of \(0.5\) at a distance of 50 tiles you would need an influence value of
\(i=0.5\cdot2,500+0.5\approx1,250\).
But all of your tiles (36 for a 6x6 sized building) have an influence effect, so in reality a value around\(\frac{2,500}{36}\approx70\) seems to be reasonable (neglecting the different distances of the tiles of the building).

The precision for influences is limited to reduce memory consumption. Therefore huge numbers (like 5000 :lol:) lead to an overflow which in return may lead to strange influence effects.
Well done :teach

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