'no two creatures more then 30' apart' simple phrase, big fight

Deset Gled said:

Luckily, the rules are completely consistent that the distance from one square to another is always counted as 5". Unless you are not using the grid system, there is no ambiguity.

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It takes a 5' reach to reach the adjacent square, true, but that's not the same thing as saying that the two squares are 5' apart.

The ambiguity exists.

I find these attempts to place a "absolutely correct" interpertation on the rules to be rather amusing.

Is it not obvious on its face than when the rules generate a discussion like this that the rule, as written, is insufficently precise?

frankthedm said:

OP was right. An adjacent creature is 5' away. If you have one square seperating you from your target, it is 10 feet away.

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This is the reasoning I would agree with also. However, it is one of those things that can vary from game to game.

Which ever way you go, just make certain you stay consistent with that in the future.

END COMMUNICATION

Artoomis said:

It takes a 5' reach to reach the adjacent square, true, but that's not the same thing as saying that the two squares are 5' apart.

The ambiguity exists.

I find these attempts to place a "absolutely correct" interpertation on the rules to be rather amusing.

Is it not obvious on its face than when the rules generate a discussion like this that the rule, as written, is insufficently precise?

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Yup, which is why after reading the Rules of the Game reference - I have changed my mind.

Skip gave an explanation of how to run this scenario and even though RotG is only above Cust Serv on my list of "precedence" it is still on the list.

So barring any other "more precedent" writing I default to that one.

irdeggman said:

Yup, which is why after reading the Rules of the Game reference - I have changed my mind.

Skip gave an explanation of how to run this scenario and even though RotG is only above Cust Serv on my list of "precedence" it is still on the list.

So barring any other "more precedent" writing I default to that one.

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Um... I don't recall Skip's "Rules of the Game" for this scenario. Which way did he rule - is "30' apart" the same as 6 unoccupied squares or 5 unoccupied squars?

edit: Found it:

"Sometimes, you can choose targets only within some sort of limited area. The rules usually use one of two different kinds of wording to indicate that. For example, the targets entry for the animal growth spell is as follows: "Up to one animal (Gargantuan or smaller) per two levels, no two of which can be more than 30 ft. apart."And the targets entry for the animal shapes spell is as follows: "Up to one willing creature per level, all within 30 ft. of each other." Usually, when the rules say things in different ways, they mean different things, but not in this case. All the targets you choose must be with the specified distance of all the other targets, and any target that is more than the specified distance from even one other target can't be selected as a target. To put it another way, imagine a sphere with a diameter (not radius) equal to the specified distance. All the targets you choose must fit within that sphere."

I wish they haad just stated "15' radius" or 20' radius." Stupid squares - hexes would make this so much easier.

Artoomis said:

To put it another way, imagine a sphere with a diameter (not radius) equal to the specified distance. All the targets you choose must fit within that sphere.

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Of course, it's possible to have three targets in a triangle, all 30 feet apart, where all three cannot fit into a 30' diameter circle.

It's a useful approximation, but it's not exactly the same thing.

-Hyp.

irdeggman said:

It doesn't say it goes out to 30 ft it says the affected targets are no more than 30 ft apart.

To positively calcualte this you need to know where to start measuring.

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Fortunately, that's easy: you start in the square of one of the creatures.

...then you start counting squares. 1, 2, 3, 4, 5, 6. At the 6th square, you stop (careful of every-other diagonal!). You've now gone 30 ft. If you've stopped short of another creature, you can't include that creature in the spell effect.

Nail said:

Fortunately, that's easy: you start in the square of one of the creatures.

...then you start counting squares. 1, 2, 3, 4, 5, 6. At the 6th square, you stop (careful of every-other diagonal!). You've now gone 30 ft. If you've stopped short of another creature, you can't include that creature in the spell effect.

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You did see where I stated that after reading the RotG article I changed my mind didn't you?

Just checking.

Hypersmurf said:

Of course, it's possible to have three targets in a triangle, all 30 feet apart, where all three cannot fit into a 30' diameter circle.

It's a useful approximation, but it's not exactly the same thing.

-Hyp.

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I don't think you can have 3 people in a triangle where they are all 30' apart and not also fit along the edge of a 30' diameter circle. That can certainly happen in D&D when the "circle" is made of squares, though.

Artoomis said:

I don't think you can have 3 people in a triangle where they are all 30' apart and not also fit along the edge of a 30' diameter circle.

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Pythagoras disagrees with you.

Let's take two people exactly 30 feet apart. They can fit on the edge of a 30' diameter circle, as long as they are exactly diametrically opposed, right?

Now let's add a third person to form an equilateral triangle. He is 30 feet from person A at a 60 degree angle, and he is 30 feet from person B at a 60 degree angle.

The perpendicular bisector of the line between A and B also crosses through the centre of the circle and through person C, right? And the length of that line from centre to person C, L, is such that 15² + L² = 30². So L² = 675, and L ≈ 26. 26 is much bigger than 15.

-Hyp.

Hypersmurf said:

Pythagoras disagrees with you.

Let's take two people exactly 30 feet apart. They can fit on the edge of a 30' diameter circle, as long as they are exactly diametrically opposed, right?

Now let's add a third person to form an equilateral triangle. He is 30 feet from person A at a 60 degree angle, and he is 30 feet from person B at a 60 degree angle.

The perpendicular bisector of the line between A and B also crosses through the centre of the circle and through person C, right? And the length of that line from centre to person C, L, is such that 15² + L² = 30². So L² = 675, and L ≈ 26. 26 is much bigger than 15.

-Hyp.

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Whoops: Right you are. The actual minimum radius is the square root of (15^2 + 13^2) (I'm pretty sure), so that's about 19.6 - so I guess using a 20' radius area of effect template is close enough. (That, btw, is the distance to the center point of that 30' on-a-side equalateral triangle mentioned above)