Fix Advanced Calculation

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4 comments

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    Matt Weigel

    Why don't you just use the area of the space to get the square footage of crack isolation membrane you need given you sell it by the SF? Maybe I'm missing something, but I generally add my material (in this case tile), to a drawn area, then as an addon to that drawn area, I will add my crack isolation membrane (i.e ditra mat, stratamat etc.) If you want to order more than just the net area in isolation membrane then just make an advanced formula on the isolation membrane product that takes the netarea and multiplies by 1.05 for 5% or whatever you want.

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    Jerry Johnson

    Matt, 

    thank you for your comment!  If I were selling ditra I would use the method you propose, but I'm talking about a membrane that is used strictly to isolate saw cuts and/or cracks - not full coverage.

    I'm using a line (point A to point B), and therefore it has no SF, only a linear foot.  That is why I need a calculation.  

    I would simply use a room add-on, except that I only need crack iso at the concrete saw cuts and not the entire room area.  If I want to use a room add-on I would have to figure it manually per room, which would be a much worse scenario.

    In summary, I sell the membrane by the SF and the labor by the LF.  Both are based on the LF of saw cuts.  As such, the SF needed will not equal the sf area of the room.

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    Matt Weigel

    Ahhh, that makes more sense, thanks for the clarification, Ill keep thinking if there might be another way to make that work

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    Jerry Johnson

    Thanks for pressing for the clarification Matt!  I didn't  realized I had overlooked that key point :-)

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