Area of Effect/Size Printable Templates?

frankthedm · Mar 26, 2007

I currenly do my grids / templates on a grid where each box is 59 pixels across and the lines are one pixel themselves. One or two were done by merging 4 squares into one, but those are larger than I like working with.

frankthedm · Mar 26, 2007

These were done on 96x96 pixels to the inch grids

Ogrork the Mighty · Mar 26, 2007

Awesome! Thanks!

frankthedm · Mar 27, 2007

carborundum said:
I'm inclined to say the first one. The vertical axis mirrors the horizontal one, unlike #3, and the transitions seem fair.

Seems fair? Fair often takes a back seat to rules. Have we forgotton how the counting squares rather than measuring steals area from a 20' radius? That theft does have to be replicated in 3d to achieve a proper result. If it seems "fair" too much area has been given.

frankthedm · Mar 28, 2007

Here is an example of grid based radii 5' to 40' across. Also included is what the rounded radii look like in comparison.

Nail · Mar 28, 2007

Huh.

I never noticed how much area you lose from a radius by using the "count the squares" method.

Huh.

BTW: I used the Steel Squire templates last night in game -- I even modified my bad guy a bit so that I could use them all! (20'r., 10'r., 30' diagonal cone, 30' straight cone) Great fun, and definitely worth it.

carborundum · Mar 29, 2007

frankthedm said:
Seems fair? Fair often takes a back seat to rules. Have we forgotten how the counting squares rather than measuring steals area from a 20' radius? That theft does have to be replicated in 3d to achieve a proper result. If it seems "fair" too much area has been given.

Ah, I forgot the spurious mathematical justification! Humble apologies!

A quick 4/3*PI*r^3 (with r=4 square-lengths) gives us a volume of 268 cubes.
Counting cubes in the templates gives 208 cubes, 168 cubes and either 228 or 232 cubes (depending on the corners).

Initial evidence would point to the third one being the most reasonable.

However...

When using a 20' radius template and counting squares, we are told to use 40 squares, while the real area is about 50. One could argue that 80% coverage is therefore demanded by the rules, and choose the template giving the closest to (268*0.8) 214 cubes. Number one!

I await your critique of this spurious justification with a quivering bladder.

frankthedm · Mar 29, 2007

carborundum said:
Ah, I forgot the spurious mathematical justification! Humble apologies!

A quick 4/3*PI*r^3 (with r=4 square-lengths) gives us a volume of 268 cubes.
Counting cubes in the templates gives 208 cubes, 168 cubes and either 228 or 232 cubes (depending on the corners).

Initial evidence would point to the third one being the most reasonable.

However...

When using a 20' radius template and counting squares, we are told to use 40 squares, while the real area is about 50. One could argue that 80% coverage is therefore demanded by the rules, and choose the template giving the closest to (268*0.8) 214 cubes. Number one!

I await your critique of this spurious justification with a quivering bladder.

Thanks for the math. It is not my strong point.

One seems about right. As I said before the rules are not kind to area effects. One does not get measure area to determine what is effected. One has to count squares. We already have a three for two rule when going diagonally in 2D. For 3D, we have to first figure out what standard of square counting to use. The amount that cliped may wind up increasing once a third dimension is introduced. Once that standard is made, the it has to be followed, regardless of how badly it clips the area.

>>>>>>>>>>>

Lets say going from the two corners farest apart in the cube costs 2 squares, which of the three in this post match that best?

carborundum · Mar 29, 2007

Did you mean how does the three for two work out numbers-wise in diagonal steps?
If you take a 2x2, Mr. Pythagoras says the diagonal is 2.8 (square root of 8). Three for two isn't that harsh really.
For a 2x2x2 cube, the longest diagonal has one side 2 long and one 2.8. This works out at about 3.5 (3.46) - what would you call that? A cost of seven for four diagonal 3D hops, I guess.

Sorry, not quite sure what you mean with the "two furthest" thing, Mr. Frank. I'm glad you liked the argument though

Pyrex · Mar 29, 2007

frankthedm said:
Thanks for the math. It is not my strong point.

One seems about right.

One certianly seems to be the most intuitive 3d representation of the 2d template even if it's not the most accurate model of a 20' radius in 5' cubes.