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STRUCTURAL RENOVATION, ROOFING, DEMOLITION, etc
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  #1  
Old 3rd Dec 2007, 10:11 AM
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Default Setting out and stringlines.

Just had a query regarding setting out. If you are setting out from say, an existing brick wall, and you measure out and make a stringline at the end of your extension parallel to the wall, when it comes to setting out the perpendicular stringlines at the edges you have to be careful to avoid creating a parallelogram instead of a square correct?

For example, if you set out one stringline angling to the right away from the house, and another angling away to the left, then when you come to measuring the diagonals you won't have a square when you make them the same...you will have a parallelogram correct? So to avoid this potential problem is it best to angle the perpendicular stringlines by eye in the same direction away from the house THEN square them up by measuring the diagonals?

As a secondary question, how do you get the stringline running parallel to the house to be perfectly parallel? I was working with someone and from memory we just measured out the distance from the existing bearers, drove stakes in the ground then set up hurdles, levelled them with a laser level and did the same on the other side. Then we ran a stringline between the hurdles parallel to the house. Is that correct?
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  #2  
Old 3rd Dec 2007, 10:26 AM
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CS,
if you can run a stringline along one of your end walls (one that's square to the one your deck will be parrallel to) then you can simply measure off this. If this is not possible then you'll need to use pythagorus (3,4,5) to square your initial line.

For your parrallel line to be strictly parrallel you'll need to make your measurements square to the building, pythagorus to the rescue again. I hated maths at school and had no idea I'd end up using it so often in daily life.
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  #3  
Old 3rd Dec 2007, 10:52 AM
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Thanks, I'd thought about using pythagoras for the second part of my question, and yeah that would get you a square line off the wall to then measure off to your hurdles and run your parallel line from (or yeah running it along an existing wall a way then out to your extension length).

However does anyone have any idea about the first part of my question, ie avoiding a parallelogram?
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Old 3rd Dec 2007, 10:59 AM
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Just remember that most houses are out of square to start with.
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  #5  
Old 3rd Dec 2007, 02:29 PM
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To be picky... squares and rectangles are parallelograms.

I assume you mean avoiding rhomboids (squashed rect's) & trapezoids (tapered rect's) but if you use Pythag to lay out both end lines before measuring off for your hurdles and ensure both diagonals are the same length before placing the hurdles, then out-of-square won't be a problem.

Just take your time and get it right.

Of course, if one end wall is going to be a continuation off an existing corner, you're often better off running a string line along that wall, even if it means that the result isn't properly squared. As Bricks said, many houses aren't squared and if you build an extension "properly" you end up with a dogleg in the wall, which looks ugly. A case of reality versus theory, I'm afraid.
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Old 3rd Dec 2007, 02:40 PM
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Technically a square is a rhombus too But a rectangle isn't...

But actually what he's talking about avoiding is a parallelogram that isn't square - the opposite sides are parallel. (A parallelogram is only a rhombus if all four sides are of equal length).

But all the jargon aside, perhaps if you post a picture of what you have now and what you're trying to lay out, someone might be able to give you a step by step on how to get there.
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Old 3rd Dec 2007, 02:54 PM
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A rhombus is a squashed square.

A rhomboid is a squashed rectangle.
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Old 3rd Dec 2007, 03:09 PM
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Technically, rhomboid is not all encompassing in this situation, because it does not cover square extensions, where as parallelogram does

A rhombus is a quadrilateral with four equal and parallel sides - a square is a rhombus with right angles, so it doesn't have to be squashed to be a rhombus, so we still have the same problem as you are trying to solve with parallelogram. In fact I don't know if there is a term for a quadrilateral with parallel sides and oblique angles - except rhomboid, which, as I said, doesn't cover parallelograms with 4 equal sides.

Damn I hated geometry at school, why are you making me remember all this?
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Old 3rd Dec 2007, 03:10 PM
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Ok guys, I've posted a couple of pics here to help outline the question. In the first one, assuming that both those stringlines were at the same angle out from the wall then when I measure the diagonals they will be the same measurement, but it's obviously not a square.

If however, I set out the stringlines like the second picture, then I can bring them into a square by checking the diagonals and adjusting accordingly...this wouldn't happen in the first scenario. Considering that the initial measurement out from the wall is often just with a tape and a stake, is it a good idea to bear that in mind?
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Old 3rd Dec 2007, 03:25 PM
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In that case, you have a fixed reference point, which is the rear wall of the house, so I would just use 3,4,5 to set both lines and then check the distance at both ends to make sure they're parallel.
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Old 3rd Dec 2007, 04:17 PM
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Quote:
Originally Posted by silentC View Post
A rhombus is a quadrilateral with four equal and parallel sides
If we're being technical...

Can you have a 2 dimensional geometric shape with 4 parallel sides? Rhombus and rhomboid shapes have opposite sides parallel.

A rhomboid is an oblique-angled parallelogram with only the opposite sides equal in length (and the opposite angles equal). A rhombus is an 'oblique-angled' equilateral parallelogram, so a square is not really a rhombus. And a rectangle is not really rhomboid.

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Old 3rd Dec 2007, 04:28 PM
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No, a rhombus (according to my sources) is an equilateral parallelogram. The angles don't have to be oblique. So a square is a rhombus. A rectangle is not a rhomboid but it is a parallelogram.
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Old 3rd Dec 2007, 04:39 PM
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Quote:
Can you have a 2 dimensional geometric shape with 4 parallel sides?
Yeah yeah, I was taking a short cut because I assumed everyone would get that bit by now...
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Old 3rd Dec 2007, 05:23 PM
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"Of quadrilateral figures, a square is that which is both equilateral and right-angled; an oblong that which is right-angled but not equilateral; a rhombus that which is equilateral but not right-angled; and a rhomboid that which has its opposite sides and angles equal to one another but is neither equilateral nor right-angled. And let quadrilaterals other than these be called trapezia."

Euclid - Elements, Book I, Definition 22



Quote:
Originally Posted by ComboSquare View Post
Ok guys, I've posted a couple of pics here to help outline the question. In the first one, assuming that both those stringlines were at the same angle out from the wall then when I measure the diagonals they will be the same measurement, but it's obviously not a square.

If however, I set out the stringlines like the second picture, then I can bring them into a square by checking the diagonals and adjusting accordingly...this wouldn't happen in the first scenario. Considering that the initial measurement out from the wall is often just with a tape and a stake, is it a good idea to bear that in mind?
If you set out the end lines independantly using 3:4:5 then they will be square to the wall you're setting them out from, assuming that wall to be straight and not curved in the first place. Oh... and that you haven't made a mistake in calculating the 3:4:5 or are using a "stretchy" tape measure, of course!

It rather makes your problem moot.
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Old 3rd Dec 2007, 05:42 PM
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Google 'square rhombus' and find me a reference (other than something that is over 2000 years old) and I'll believe you
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