Calculating Linear Feet to Saw Cut

[HardRockMike]

Im trying to find a formula or way to enter dimensions of trenches (rectangles) or squares that we bid to concrete saw cut.
then calculate how much linear footage there is to saw cut.

Example:
20' x 1' trench saw cut into 2' x 1' pieces. (that's cutting perimeter and making cross cuts every 1')

or

25' x 25' area cut into 2' x 1' pieces.

any help would be great!

thank you

 

[Webbarr]

Hi HardRockMike, are you able to post any raw data that you may already have so I can get a better understanding?

 

[shg]

How accurate are you trying to be? You can't cut a 25 x 25 block into 2 x 1 blocks.

 

[HardRockMike]

ok, im working on a bid right now, this is some of what I have.

26' x 12' area
44' x 7' area
16' x 18' area
32' x 15'

If I Work out my example from above of a 20' x 1' trench my total LF to saw is 51 LF
2) 20 cuts the length of trench
11) 1' cuts for cross cuts
=51'

or perimeter sawing is 42'
and your 9 cross cuts every 2' = 9'
total 51' of sawing

 

[HardRockMike]

I understand that, for those we always bump it up a cut or 'add an extra cut'
Id rather be high on my total then underbid myself on the work we actually do.

on that note,

I guess would there be an easy way to interchange the size pieces I cut them into.

some customers like 4' x 4' pieces because they can get heavy equipment in to lift them.

I say 2' x 1' just because its our 'normal' pieces if we hand remove them

 

[shg]

Then maybe ...

[Table="width:, class:grid"][tr][td]Row\Col[/td][td]

A​

[/td][td]

B​

[/td][td]

C​

[/td][td]

D​

[/td][td]

E​

[/td][td]

F​

[/td][/tr][tr][td]

1​

[/td][td]

Length​

[/td][td]

Width​

[/td][td]

Length​

[/td][td]

Width​

[/td][td]

​

[/td][td]

​

[/td][/tr]
[tr][td]

2​

[/td][td]

20​

[/td][td]

1​

[/td][td]

2​

[/td][td]

1​

[/td][td]

51​

[/td][td]E2: =CEILING(A2/C2, 1) * CEILING(B2/D2, 1) * (C2+D2) + A2 + B2[/td][/tr]
[tr][td]

3​

[/td][td]

25​

[/td][td]

25​

[/td][td]

2​

[/td][td]

1​

[/td][td]

1025​

[/td][td][/td][/tr]
[tr][td]

4​

[/td][td]

25​

[/td][td]

25​

[/td][td]

4​

[/td][td]

4​

[/td][td]

442​

[/td][td][/td][/tr]
[/table]

 

[HardRockMike]

hmmm interesting. this is awesome.

how close will it work with something
like a

44' x 7' area cut into 5' x 5' pieces. I know this is an odd size for pieces on this trench but its what they want (???)

either way this is great stuff, thanks. keep them coming

 

[HardRockMike]

If I use that and do 100' x 10' trench into 5' x 5' pieces, that formula comes up short 100 LF
its telling me 410 and I think it should be 510

3) 100' cuts
21) 10' cuts

 

[shg]

If I use that and do 100' x 10' trench into 5' x 5' pieces, that formula comes up short 100 LF
its telling me 410 and I think it should be 510

[Table="width:, class:grid"][tr][td]Row\Col[/td][td]

A​

[/td][td]

B​

[/td][td]

C​

[/td][td]

D​

[/td][td]

E​

[/td][td]

F​

[/td][/tr][tr][td]

1​

[/td][td]

Length​

[/td][td]

Width​

[/td][td]

Length​

[/td][td]

Width​

[/td][td]

​

[/td][td]

​

[/td][/tr]
[tr][td]

2​

[/td][td]

100​

[/td][td]

10​

[/td][td]

5​

[/td][td]

5​

[/td][td]

510​

[/td][td]E2: =CEILING(A2/C2, 1) * CEILING(B2/D2, 1) * (C2+D2) + A2 + B2[/td][/tr]
[/table]

 

[HardRockMike]

well hells bells why didn't it work for me