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Any specific formula
show user profile  sheheryar_noor
Hi guys, i was bridging a cylinder with box. 24 Sided cylinder required 6x6 segments box, so each of the Cylinder's edge would bridge with single edge of box, which means each edge gets a single edge.
Same 64 sided cylinder required 16x16 segments box to bridge with.
But i got the correct number of box edges after many tries,so i wanna ask is there any simple formula for that if i have 40 sided cylinders how many segments would i required in box?
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4/3/2014 8:51:13 PM (last edit: 4/3/2014 8:51:13 PM)
show user profile  Mr_Stabby
boxes have 3 dimensions of segments, you need any 2 of these dimensional segments to add into the number of cylinder segments / 2 since a box segment goes across to the other side.
so 6x6x1 = 12+12=24
16x16x1 = 32+32 = 64

4x5x6 could be 8+10 = 18 or 10+12 = 22 or 8+12 = 20
and so on

not really a formula.. just logic

read 415 times
4/3/2014 9:00:19 PM (last edit: 4/3/2014 9:01:30 PM)
show user profile  sheheryar_noor
:) This"4x5x6 could be 8+10 = 18 or 10+12 = 22 or 8+12 = 20" just goes above my head :D. But what i got from your logic is that it should be one third of the cylinder's edges
E.g we have 40 sided cylinder so half of 40 is 20 and then half of 20 is 10 so we would need 10x10x1 .Am i right? and i didn't mention the third dimension because i thought not mentioning means default 1. But still i have confusion what if we have 29 sided cylinders. i kept trying but every time there was an edge bridged with 2 edges so i had to delete it.
Hehe can you simplify it ?:P what if we have 27 29 31 32 33 sided cylinders. How many segments we would required for box?
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4/3/2014 10:12:56 PM (last edit: 4/3/2014 10:12:56 PM)
show user profile  Mr.Burns
Ideally you'd take an evenly sided box, in that case it'd be 1/4 of the cylinder's sides. This only works if the cylinder has n * 4 sides of course. The logic behind this is that the box has 4 sides, and all 4 combined need to have as many segments as the cylinder.

You won't be able to get a box with an odd number of sides though.
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4/3/2014 10:50:19 PM (last edit: 4/3/2014 10:50:19 PM)
show user profile  Mr_Stabby
>> just goes above my head
the viability of these options depends on which axis you're trying to bridge

>> and then half of 20 is 10 so we would need 10x10x1 .Am i right?
yes, right. But also it could be 5x15 or 3x17

if its odd sided cylinder then you're out of luck, since each segment adds 2 sides only even number is possible

either way, its best practice to model stuff with an even number of edges in the longitudinal direction, this way no matter where you're model ends up, you're always guaranteed to be able to close it up with quads. For instance if you're modeling an arm and you have 7 edges going towards the hand, you'll be forced to stick a triangle in there somewhere.

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4/3/2014 10:56:51 PM (last edit: 4/3/2014 11:13:09 PM)
show user profile  sheheryar_noor
Gotcha :). And for even numbers i choose the number which is divisible by 4 and that's help me a lot :)
read 354 times
4/4/2014 12:35:44 PM (last edit: 4/4/2014 12:35:44 PM)
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