Help with "want to try DIY shelf"

Defunct_Mather

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Jun 21, 2018
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Hello, I want to make a shelf. My initial idea of the end product is attached. I have
Horizontal pieces = 22" x 3 (shelves) x 2 (front/back) + 11" x 3 x 2 = 198"
Vertical pieces = 42" x 4 (sides) = 168"
Sum of pieces = 366"

The material for the rods that I want I know come in 72" (Online) but I think I had seen 60" (In-Store).
That would mean I would need: 366" / 72" = 5 1/12 or 366" / 60" = 6 1/10 number of pieces.
Obviously, 1/12 and 1/10 means absolutely nothing in this real world example.
So when I go into the store (need to go in so person can cut for me), I would need to tell him/her something like "cut this one into two 22" pieces and that one into..."

I would need to know what is the most efficient way to tell them to cut the rods without much waste; I would need someone to do the math for me since I don't know what to do. Heck, there could even be no issue here and I just say cut them and still just use 6 or 7 pieces, or I would just have to brute force this problem trying out alot of combinations. Since I want less waste, the bottom portions have a tolerance range

Seeming, how I don't remember if the size in store differs online, could any volunteer help with the different scenarios.

S1:
rods are 72"

S2:
rods are 60"

Bonus (for me :D):
I want to make the Basin (1st shelf) first. This would test out, could the shelving hold my aquarium before I make the full thing and realize it's useless. What would be the minimum dimensions to create just that one with regards to future pieces?
i.e. I have 2 rods. (1) cut into 2x22" and 2x11" leaving 6". Legs will be 6" giving (2) cut into 3x6", 2x20" and 1x14", if 72".

Thank you for any assistance you guys can do.
 

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Hello, I want to make a shelf. My initial idea of the end product is attached. I have
Horizontal pieces = 22" x 3 (shelves) x 2 (front/back) + 11" x 3 x 2 = 198"
Vertical pieces = 42" x 4 (sides) = 168"
Sum of pieces = 366"

The material for the rods that I want I know come in 72" (Online) but I think I had seen 60" (In-Store).
That would mean I would need: 366" / 72" = 5 1/12 or 366" / 60" = 6 1/10 number of pieces.
Obviously, 1/12 and 1/10 means absolutely nothing in this real world example.
So when I go into the store (need to go in so person can cut for me), I would need to tell him/her something like "cut this one into two 22" pieces and that one into..."

I would need to know what is the most efficient way to tell them to cut the rods without much waste; I would need someone to do the math for me since I don't know what to do. Heck, there could even be no issue here and I just say cut them and still just use 6 or 7 pieces, or I would just have to brute force this problem trying out alot of combinations. Since I want less waste, the bottom portions have a tolerance range

Seeming, how I don't remember if the size in store differs online, could any volunteer help with the different scenarios.

S1:
rods are 72"

S2:
rods are 60"

Bonus (for me :D):
I want to make the Basin (1st shelf) first. This would test out, could the shelving hold my aquarium before I make the full thing and realize it's useless. What would be the minimum dimensions to create just that one with regards to future pieces?
i.e. I have 2 rods. (1) cut into 2x22" and 2x11" leaving 6". Legs will be 6" giving (2) cut into 3x6", 2x20" and 1x14", if 72".

You haven't actually said how you are making the shelves out of rods (are they just the perimeter, and some other material fills in the middle?), or how their thickness fits in here. I'll just assume you are right that you need

4x42"
6x22"
6x11"

To cut these from the fewest possible 72" rods is not a lot of math; you've already done it in your addendum. If necessary you could just make a paper model of what you need and try fitting the pieces together into rods of the right length, with no real math at all.

But clearly each 42" piece has to come from a separate rod; that uses 4 rods and leaves 4 30" pieces. The best you can do with those is to cut a 22" piece from each, leaving 8" scrap. Then you need 2 more 22" and 6 11", so you cut 2x22" and 2x11" from a fifth rod, using 66" and leaving 6" scrap. Then you'll need a 6th rod, from which you cut the rest:

#1: 42+22+(8)
#2: 42+22+(8)
#3: 42+22+(8)
#4: 42+22+(8)
#5: 22+22+11+11+(6)
#6: 11+11+11+11+(28)

I don't see any way to do better.

The same sort of reasoning would work for the 60" rods.
 
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