Ranged combat, distance determinations?

[Gwiber]

Recently my party ran into a org group of size, the four of us were more or less over powered by the situation, the decision was made to climb the trees around us as the ogres came to physical range, they had no ranged weapons. We possessed bows and spells.

The question was broached, which I am looking for an answer for: How do we calculate distance when diagonal ranges are involved?

[Hakka Tsadok]

Pythagorean Theorem!

Distance = square root of ((height squared) + (length squared)).

And you thought you would never use this in real life.

[Gwiber]

LOL Well, yeah! But is that how the rules for the Roleplaying game calculate it?

[Isil-zha]

Gwiber, yes, at least roughly

That is the reason why every second diagonal counts twice. This is a factor of 1.5 on average which simulates the sqrt(2) factor you usually get for a square's diagonal.

[Kayerloth]

Or in other words they do it the same way you would on the ground for horizontal difference. I'll often just imagine it tipped upright from drawn out on the ground, if that makes sense.

[Thomas Long 175]

Hakka Tsadok wrote:

Pythagorean Theorem!

Distance = square root of ((height squared) + (length squared)).

And you thought you would never use this in real life.

O.o you're kidding right? Pythagorean theorum is used to simulate every force and distance in reality.

But yes the game goes 1 for first diagonal, 2 for following, repeat. square root of 2 for a diagonal would be 1.41 approximately.

[Gwiber]

So?

Count from the base of shooter to base of tree and then count how far UP the tree (in this case)?

20 feet from tree and then 20 foot up the tree in the games rules would be.. uh. what? Without Pythagorus?

(With - 28.8. 30feet?)

[Mapleswitch]

You would count from the corner of your foot to the corner of their face.

[Maezer]

Gwiber wrote:

So?

Count from the base of shooter to base of tree and then count how far UP the tree (in this case)?

You would count from the closest corner of the attack to the closest square of the target.

From what you described to me. The target is is sqaure (4,4) from the attacker at (-1,0),(0,0),(-1,1),(0,1). Ignoring the third axis because you weren't using it.

5' distance would hit: square (1,2)
15' distance would hit: square (2,3)
20' distance would hit: square (3,4)
25' distance would hit: square (4,4)

[RedDogMT]

If you need to, do a quick sketch of the scene from a side view. Then, count the squares the way you normally would for determining range.

[Speaker for the Dead]

Thomas Long 175 wrote:

Hakka Tsadok wrote:

Pythagorean Theorem!

Distance = square root of ((height squared) + (length squared)).

And you thought you would never use this in real life.

O.o you're kidding right? Pythagorean theorum is used to simulate every force and distance in reality.

But yes the game goes 1 for first diagonal, 2 for following, repeat. square root of 2 for a diagonal would be 1.41 approximately.

For what it's worth, it's said that Pythagoras worked out his theorem using the tiles in his bath. Maybe he was a gamer?

[Krodjin]

Don't forget to add your +1 to hit on account that you have higher ground.

Its already been said but I will reiterate - if you don't want to do the math or you just want the "game" distance draw it out (like you are looking at the scenario from the side, rather than above) and just count the squares, using the "every 2nd diagonal counts as 2 squares" rule.
Done & done.

[RedDogMT]

Krodjin wrote:

Don't forget to add your +1 to hit on account that you have higher ground.

+1 to hit for higher ground only applies to melee.

[Gauss]

Here is an image for top down vs side view.

The archer is 30feet above the floor and 25feet above the Ogres.
He is 25feet from Ogre1 and 35feet from Ogre2.

How I set this image up:
First, I count the range in the standard top down view.
Ogre1 is 25feet from the Archer.
Ogre2 is 35feet from the Archer.

Next, I put the ogres at that distance on the X axis from the Archer in the side view.

Now I can count the distance from Archer to Ogres. It is 35 feet to Ogre1 and 45feet to Ogre2.

Lets check via Pythagoras
Ogre1: (25^2+25^2)^0.5 = 35.355 (round down to 35)
Ogre2: (35^2+25^2)^0.5 = 43.011 (round up to 45)

Steps to do at your table:
1) Count top down distance to your target.
2) Quickly sketch out the distance height vs distance to target. Count squares.

Or
1) Count top down distance to your target.
2) Plug (height^2+distance^2)^0.5 in a calculator.

- Gauss