I made this little chart for myself and thought I’d share…
This chart shows the lengths of diagonally-chiseled walls and the nearest achievable lengths of corresponding orthogonal walls, should you want them as close to congruent as possible. The ORTHO column shows the length (in blocks/meters) of a straight wall, then the DIAG column is the amount of blocks you would have to place diagonally for [almost] the same length.
The ortho lengths highlighted in green are all within 1/10 of a block using a slope chisel, and 1/20 of a block using a bevel chisel. Conversely, everything highlighted orange is within those same fractions of a half-block. DEVIATION shows the difference between the EXACT length of diagonal walls, as found by the Pythagorean Theorem, and the exact length of orthogonal walls we are able to build.
Since beveled blocks slopes are the same as those of sloped blocks, only scooted a half block, they can be twice as precise and give us far more options to achieve near-congruent walls. The .5’s in the DIAG column are when you have a full block with just the corner beveled out, while whole numbers are with the addition of the filler corner. Just remember to count that half of the orthogonal block when measuring for a beveled wall.
This is in a spreadsheet I made in Excel. I could convert it to Google Sheets or whatever would be helpful if anyone wants it. I can also reverse engineer the formulas to show how many diagonal blocks would be needed per length of orthogonal blocks, it would just be a longer list with less precision.
Feel free to message me if you’d like to know a specific length but don’t feel like using the spreadsheet!
Darn it my brain is too fried buying spray cans in every color group for me to handle this :o
Hah I’m so bad at explaining stuff. A 41-block wide straight wall is almost exactly as long as a 29-block wide copper chiseled wall. That’s the most accurate one. If you start there the others should make sense
I think a picture would make wonders here
" A picture is worth a thousand words "
Yikes. You took that build seriously lol.
Personally I just kept chucking down blocks and chiselling them until I had a diagonal angled gate (open canal lock was the goal)
Total yolo and winging it does lead to some grrrrrr smash smash moments though
Does this help?
With some chisel:
Bevel is similar, but moved half a block over.
I thought, if you had 90 degree corners, it was 2:1 for diagonals. As in a 2x2 square is the same with as 1x1 diaganal. Not at comp atm, but will share a screenshot when on…
Edit: also if a(squared) + b(squared) = c(squared), and a and b are the same, 2:1 makes sense…
That wouldn’t be possible or a and b would have to be a straight line to be as long as the hypotenuse. The ratio is the 1.414 from the first line.
I forgot to take pics last night so thanks @Rydralain. They don’t really look even to my eyes but maybe that’s an illusion? I’ve done 7:5 a few times in builds.
wouldnt that mean… u would require 1.414 more plots to achieve the same amount of space?
If 1(squared) + 1(squared) = 2(squared) then the answer for “c” is the square root of 2.
Bro I had flashbacks to highschool math class reading this…
Im still confused, how is 0.071067812 deviation a difference in with of 2 blocks…
Well i am terrible at maths… They are defiantly not even, the 5 diagonal are two blocks wider…
Sorry if i’m being really stupid, but I honestly cant work this one out and it is bugging me.
Here is what i meant by 2:1:
You see the volume is 2:1 for the same with, as well as the number of sides (because they are squares):
From my understanding:
X(squared) + X(squared) = Y(squared) =
2 x X(squared) = Y(squared) =
X(squared) = Y(squared) / 2
I feel like Pythagoras is confusing the situation, looking at the screenshots all you need is the new Y to equal the old X…
I mean all Pythagoras is really saying here is that the diagonal of a square is longer than its sides.
edit: you can take a ruler to the screenshots provided by Rydralain (or your own for that matter) and check how long the sides are. @Samski
edit2: be sure to check the SIDES though, because the width of the chiseled square is the diagonal of the unchiseld one
I forgot to lay this out again. Those 2:1 sides aren’t the same length though. Do a 20 ortho x 10 diagonal wall, 30x15, the difference will keep growing. The last line you did is redundant because you’re solving for y. If you just plug numbers in, say the ortho sides are 5:
5sq + 5sq = Ysq
25 + 25 = Ysq
Square root/ 50 = Y
Y is close to 49 which is the square of 7, so Y will be slightly longer.
If Y were 10, then it would be 25+25=100. Imagine you have two 5” sticks and one 10”. Only way for all three sticks to meet is if the 5” ones are laid flat over the 10”. Hypotenuse always has to be shorter than the other sides (combined).
I first did this when I was making an octagonal build with the first wall at 30 blocks and was guessing on the diagonal sides, but I could tell it was uneven. So I ran the numbers and ended up changing it to 34. It appeared perfectly even after that.
Hey, this is useful. Thanks.
Thanks man! I’m about to make a bunch of chandeliers so I’ll be using the bevel chart for some of them. I added that part yesterday, hadn’t opened the file since I made the slope one last year.
Now A understand, you want the length of A and B to the the same, whilst at a diagonal, not the same with in overall object shape.