Test Release v3.2.0

[El Condoro]

Someone else has mentioned it, but the Items display seems to interact with all open modules, which makes it practically unworkable. If the modules are closed (DD PHB Deluxe, DD SRD Magical Items) first the Items dialog box opens quickly and can be worked with.

Edit: The NPC box does, too (DD MM Monster Manual and any other NPC modules) but in my tests this hasn't caused any performance issues.

[El Condoro]

Being picky here and it's not a big deal but, although the new 5E cone is great but it's not quite right. In the image you can see the length of the sides is actually 45 ft. but the cone is showing 50 ft. which is the distance from the origin to the arc of the cone. The triangle formed is therefore not quite an equilateral one, as per the 5E rules.
Attachment 14697

[damned]

Being picky here and it's not a big deal but, although the new 5E cone is great but it's not quite right. In the image you can see the length of the sides is actually 45 ft. but the cone is showing 50 ft. which is the distance from the origin to the arc of the cone. The triangle formed is therefore not quite an equilateral one, as per the 5E rules.
Attachment 14697

Being a little pickier again - its not quite the 5e cone but not for the reason you say... if you measure the distance from bottom left end of the cone to the top left end of the cone it is narrower than it is long.... is the 5e cone meant to be a triangle rather than a cone?

[Zacchaeus]

A cone’s width at a given point along its length is equal to that point’s distance from the point of origin. So it can't be an equilateral triangle; unless I've forgotten everything about Higher Maths. Which is certainly possible given the length of time since I sat that particular exam.

If the cone is 15' measured from the point of origin to it's end then the sides will measure 17' - I believe.

[damned]

A cone’s width at a given point along its length is equal to that point’s distance from the point of origin. So it can't be an equilateral triangle; unless I've forgotten everything about Higher Maths. Which is certainly possible given the length of time since I sat that particular exam.

If the cone is 15' measured from the point of origin to it's end then the sides will measure 17' - I believe.

Thats different to spites description...

As mentioned numerous times ine very thread on this topic, the FG cone is not the correct cone for 5th edition. 5th edition cone is equal in diameter to the length from the beginning point.
IE: a 10foot cone has the widest point 10foot wide, this is not the case with the FG cone which maintains a 90degree angle.

this describes equal side lengths and equal distance between those points. what it doesnt say is curved or straight end... triangle (equilateral) or cone?

[El Condoro]

The language is a bit cumbersome, in my opinion, but if the width = the length then it must form an equilateral triangle with an arc. Actually, if we look at the cone drawn on p. 204 of the PHB and imagine it side-on (2D), it would look like an equiateral triangle without an arc at all.
Attachment 14698

[dulux-oz]

No, its not an equilateral triangle (although its about as close as you can come to drawing one without a protractor or a compass).

An equilateral triangle is one where all three sides are the same length (say, 2 inches) and all three (interior) angels are 60 degrees. If you draw a line from an apex to the middle of the opposite side (known as bisecting the angle) you end up with a triangle with angles of 30, 60 and 90 degrees, and sides of 2 inches (the base or hypotenuse), 1 inch (the shortest side, half the old triangle's side) and the new side which is the line that we bisected the angle with. This new line is NOT 2 inches long (its (the square root of 3) inches long) and so the triangle described as having its width at a given point along its length is equal to that point’s distance from the point of origin is NOT an equilateral triangle.

For the record, the angle at the origin or apex of such a triangle can be worked out via the following formula: twice the inverse tan of (opposite length/adjacent length) = 53.13 degrees (approx)

With working: opposite length = 1 inch, adjacent length = 2 inches.
1 divided by 2 = 0.5
inverse tan of 0.5 = 26.565 (approx)
twice this result = 53.13 (approx)

See, being a Mathematician comes in handy sometimes

[Zacchaeus]

I was fine until you got to inverse of tan

I though the inverse of tan was 'pallid', like us Scotsman this summer

[dulux-oz]

I thought you Scotsmen shriveled up in the sun, like vampires?! (And that's why you guys make such good Whiskey - it keeps you "hydrated".)

[Zacchaeus]

We do indeed shrivel in the sun, but a little bit isn't too harmful. And since we are being pedantic Whiskey is from the States; we produce the superior Whisky