The new netting design. Your help needed...

[OllieMartin]

I was going to suggest using a drawstring all round the bottom rather than elastic, but you could use big pegs instead, I think.

So, you have a long piece of voile, stitch top & bottom to each other to form a tube (circumference of tube = outline of bed) then pop the tube over the frame, gather the top together (tie with string or elastic band) and peg the bottom to the bed. V basic sowing needed - 1 seam - and if you really are that hopeless with a needle & thread you could velcro that.

You're a useful person to be friends with!
Definately a workable way of doing it. The voile sheets are a maximum of 1m in width and it would be a minimum of 2.2m from the base to the top, so I'll either need to sew them (or staple them) together or find some alternative.

[OllieMartin]

Okay, maths time:

I need to calculate how long each of the ribs needs to be.
This should all be calculatable from the bed dimensions of 2mx2m (axb). And the assumption that the diagonal ribs are semi-circles.

The diagonals are fairly easy to calculate:
Using Pythagoras:
Diameter=sqrt(a^2+b^2)=sqrt(4+4)=2.82m
Radius=2.82/2=1.41m
Circumference=(2 x Pi x radius)/2 = Pi x 1.41 = 4.42m

However, the two ribs that extend from the centre of each edge will be eliptical. We know this because their radius measured from floor to top will be the same (1.41m), but as they span the length of a 2m bed, their radius from centre to edge will be 1m.
So the question is, "What's the circumference of an ellipse"?
I've found several different formulas, each gives a different result:

An online Ellipse Circumference Calculator says:
c=Pi(sqrt(2((a/2)^2 + (b/2)^2)))= 7.49m = 3.74m for half an ellipse

Another online Circumference of Elipse Calculator says:
c= 2 x Pi x sqrt((a^2 + b^2)/2)= 7.64m = 3.82m

Wikipedia says:
c=Pi( 3(a+b) - sqrt(10ab + 3(a^2 + b^2))=14.85m!!! (That is for a full ellipse, where as I'm only making half of one, but 7.42m still seems waaaaay too big)

Are there any maths teachers about who could confirm one of these please?

[Bigmallly]

Well Ollie, it's a long time since I was at school but I get it as 2 lengths = 3.14mts plus 2 lenghts = 4.40mts. That was using the rule that the circumfrence = 2 x pi x r (r = 2 different measurements, half of corner to corner & half of edge to edge. Am I right in thinking that the height will then be 1metre? (Well it will if you are assuming it is a semi-circle.)

[OllieMartin]

Well Ollie, it's a long time since I was at school but I get it as 2 lengths = 3.14mts plus 2 lenghts = 4.40mts. That was using the rule that the circumfrence = 2 x pi x r (r = 2 different measurements. Corner to corner & edge to edge. Am I right in thinking that the height will then be 1metre? (Well it will if you are assuming it is a semi-circle.)

Sadly not BM.
You've assumed that both corner-corner & edge-edge are semi-circular.
Using your calculations the 4.4m arc will be 1.4m off the ground and the 3.14m arc will be 1m off the ground. These both need to meet in the middle and so both need to be the same height off the ground.

The problem arises in that the edge-edge arcs are not semi-circular, they are semi-eliptical as they need to follow the gradient of the corner-corner-semi-circular arcs.
Being a visual learner I've had to draw numerous cross-sections of the design to get my head around the shapes involved. The problem now is that I can't find a definitive formula for the circumference of an ellipse given that we know the semiminor axis is 1m and the semimajor axis is 1.4m.

[Bigmallly]

Oh bugger........I forgot the 2 lengths corner to corner & edge to edge were different. So obviously the corner to corner is going to be higher........duh.........back to the drawing board.

[Bigmallly]

Is there anyway of making both axis the same length.....Would this work?

[OllieMartin]

Is there anyway of making both axis the same length.....Would this work?

If we did that we'd end up with a circular or at least octagonal base. This would either cut off a section of each of the corners of the bed or extend onto the bed next-door.
It really needs to fit a square base.

[Bigmallly]

I'm trying Ollie but it is way above my head. I hope these few links may help you.

Desert Domes - Dome Formulas

Hemispherical Dome Construction

[OllieMartin]

I'm trying Ollie but it is way above my head. I hope these few links may help you.

Desert Domes - Dome Formulas

Hemispherical Dome Construction

I appreciate the help BM.
It's getting above my head too, but my thinking is that if I can master it I'll have learnt something!
Thanks for the links. Checking them out now.

[Bigmallly]

Good luck mate, gotta get back to work. Will follow your progress.

[Hazel at the Hill]

I appreciate the help BM.
It's getting above my head too, but my thinking is that if I can master it I'll have learnt something!
Thanks for the links. Checking them out now.

If all else fails, draw each section on squared paper & count the squares!

Actually, with my method you don't have to worry about the elliptical nature of the tubing:

Take long length of voile (length of voile = total outside edge measurement of bed i.e. total of 4 sides)
The width of voile = half the length of the longest blue pipe (the diagonal one) + a bit
stitch the width edges together - you now have a tube, the circumference of which is the same size as the outline of the bed
slip tube over pipe structure
Gather the excess voile at the top in a big topknot & tie with string.
Weigh down or peg voile edges.

You may have to take a handy box and a scarf in order to make a model, as I have just done...

[OllieMartin]

I've just thought:
I should be able to get the measurements from the Google SketchUp model as it's all drawn to scale.
The ribs aren't proper smooth arcs, they're a series of straight lines, but it'll work as a good approximation.

[OllieMartin]

I've just thought:
I should be able to get the measurements from the Google SketchUp model as it's all drawn to scale.
The ribs aren't proper smooth arcs, they're a series of straight lines, but it'll work as a good approximation.

According to this method it's 3.76m although I'd expect it to be slightly longer as this will cut inside the arc slightly. So I guess I'll go with the 3.82m calculation. It's easier to cut a little off than add a little on anyway.

[Bigmallly]

Nice one Ollie, glad you got there in the end............Hope you get better veg from this.......

[yorkshire sam]

On reading some of these posts I thought I'd entered a time warp and woke on 4 months down the line. Is it April the first already?