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Steam Park» Forums » Rules

Subject: Card clarification rss

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Popov Victor
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What does "nr of squares occupied by a ride or a stand in your park"?!

A same colour ride, from what i understood from the rules can occupy max 6 squares (3+2+1), yet on the card it says that a ride can occupy 13+ squares. What am i missing?
 
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Holger Doessing
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bubu91 wrote:
What does "nr of squares occupied by a ride or a stand in your park"?!

A same colour ride, from what i understood from the rules can occupy max 6 squares (3+2+1), yet on the card it says that a ride can occupy 13+ squares. What am i missing?

It essentially means "number of squares occupied by anything (not necessarily the same ride)". If you have e.g. a 2-red, a 1+3-black and 2 stands, you have (2+1+3+2) 8 occupied spaces.
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Popov Victor
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holgerd wrote:
bubu91 wrote:
What does "nr of squares occupied by a ride or a stand in your park"?!

A same colour ride, from what i understood from the rules can occupy max 6 squares (3+2+1), yet on the card it says that a ride can occupy 13+ squares. What am i missing?

It essentially means "number of squares occupied by anything (not necessarily the same ride)". If you have e.g. a 2-red, a 1+3-black and 2 stands, you have (2+1+3+2) 8 occupied spaces.


ok thank you the wording could have been better on this card ) "occupied by ALL YOU RIDES" for instance
 
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Alessandro Pra'
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bubu91 wrote:
holgerd wrote:
bubu91 wrote:
What does "nr of squares occupied by a ride or a stand in your park"?!

A same colour ride, from what i understood from the rules can occupy max 6 squares (3+2+1), yet on the card it says that a ride can occupy 13+ squares. What am i missing?

It essentially means "number of squares occupied by anything (not necessarily the same ride)". If you have e.g. a 2-red, a 1+3-black and 2 stands, you have (2+1+3+2) 8 occupied spaces.


ok thank you the wording could have been better on this card ) "occupied by ALL YOU RIDES" for instance

That would leave stands out though! "All your rides and stands" would do.
Thanks to holgerd for giving the correct answer, anyhow
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