John DeBoo wrote:
>
> Joe wrote:
>
> > How does one go about making sure a shed is square? Is there a
> > formula like 3' down one side, 2' down the oposite side should produce
> > 5' across? Help.
> >
> > Thanks in advance.
> >
> Maybe I'm reading this wrong (and everyone else read it right) but
> if you're talking about the floor itself, measure to opposite
> corners (an 'X' pattern). If square they will both have the same
> exact measurement. As for the walls square to the floor, thats
> already been answered 3-4-5.
If opposite walls are the same length, your method will get
the walls parallel and thus right angles. It doesn't work if
one of an opposite pair is shorter, i.e. you don't have a
rectangle. You use geometry for squaring a base. I'm
curious about your "square the walls with the floor tho."
Nobody does that. You want the floor level and you want the
walls to be perpendicular. To do that you use gravity. To
get something level you use water (absent flow and wind,
water surfaces are always level)(most people use a tube
since any surfaces connected will be level) and to get walls
perpendicular you use a plumb bob. Those methods are basic
and the standard. Of course most people just use a level
which is a derivative method, i.e., the level has to be
adjusted to something else to be sure the bubbles indicate
level and perpendicular.
On Sun, 10 Aug 2003 23:55:53 GMT, "Paul" <[email protected]>
wrote:
>How about a(squared) + b(squared) = c(squared)
>where 'a' and 'b' are the sides next to the (hopefully) 90 deg corner, and
>'c' is the hypotenuse (diagonal). It works regardless of the length of
>either short side.
A Pythy response.
The old boy would be proud.
Regards, Tom
Tom Watson - Woodworker
Gulph Mills, Pennsylvania
http://users.snip.net/~tjwatson
In article <[email protected]>,
Unisaw A100 <[email protected]> wrote:
>Measure one leg over 3'.
>Measure the other leg up 4'.
>The hypotenuse will be 5'.
>
>Use the same edge of the tape on both marks when you go to
>check the 5'.
>
>You can also double/triple/quadruple this (6', 8', 10') for
>a larger building and to check that your wall isn't "running
>out".
>
>UA100
Also: 5 ft high, 12 ft wide, *should* give 13 ft on the hypotenuse.
"Joe" <[email protected]> wrote in message
news:[email protected]...
> How does one go about making sure a shed is square? Is there a
> formula like 3' down one side, 2' down the oposite side should produce
> 5' across? Help.
>
> Thanks in advance.
And who says that public schools aren't doing their job? Apparently, it's
not just a problem in the US.
If you want a right angle, if you create a triangle with sides of 3 units, 4
units, and 5 units, the angle between the 3 and 4 unit sides is a right
angle.
If you have a floor or wall laid out and want to check it for square, make
sure the diagonals are equilateral (oops...I mean the same length).
todd
How about a(squared) + b(squared) = c(squared)
where 'a' and 'b' are the sides next to the (hopefully) 90 deg corner, and
'c' is the hypotenuse (diagonal). It works regardless of the length of
either short side.
"Joe" <[email protected]> wrote in message
news:[email protected]...
> How does one go about making sure a shed is square? Is there a
> formula like 3' down one side, 2' down the oposite side should produce
> 5' across? Help.
>
> Thanks in advance.
>
On Sun, 10 Aug 2003 16:00:38 -0400, Joe <[email protected]> wrote:
Thanks for all the replies, some good ideals here.
>How does one go about making sure a shed is square? Is there a
>formula like 3' down one side, 2' down the oposite side should produce
>5' across? Help.
>
>Thanks in advance.
In rec.woodworking
[email protected] (ToolMiser) wrote:
>You were close. 3, 4, and 5.
Yep, good ol' Pythagorous. Of course, if it is squarely dimensioned,
simply measure diagonal corners across and adjust till the measurements
are equal.
Measure diagnaly.
On Sun, 10 Aug 2003 16:00:38 -0400, Joe <[email protected]> wrote:
>How does one go about making sure a shed is square? Is there a
>formula like 3' down one side, 2' down the oposite side should produce
>5' across? Help.
>
>Thanks in advance.