Filling an area with triangles to minimise space loss

[Jt2015]

As seen in the drawing attached, there are two circles, the first circle will be a container to hold water and the outer is a second layer of aluminium, in between these will be insulator.
What is the best way to work out minimum space lost by using triangles/sectors to fill this area?

Having second thoughts it may be easier to use annular sectors instead to fill the space but I'm tripping myself up with how to make them fit

 

[Merlin3189]

If the material is 80mm thick and the annulus has an 80mm gap, only thin slivers will fit.

If A is 80mm, B must be smaller, or if B is 80mm, A must be larger. The narrower the slice, the less difference there is.

If the insulating material is inflexible, I guess you just have to cut lots of thin slices.

If the material is flexible, you just bend it round to fit. It may help to do that in sections. It may help to make cuts on the inside surface - some deep, some less deep - perpendicular to the curve. The number of cuts will depend on the thickness of the blade (or the gap it creates).

I don't know about advanced insulating materials, but simple insulation usually has lots of small air pockets, because air is a poor conductor. So provided any unfilled gaps are small, I would not expect them to reduce the effectiveness of the insulation.

 

[Jt2015]

If the material is 80mm thick and the annulus has an 80mm gap, only thin slivers will fit.
View attachment 89053
If A is 80mm, B must be smaller, or if B is 80mm, A must be larger. The narrower the slice, the less difference there is.

If the insulating material is inflexible, I guess you just have to cut lots of thin slices.

If the material is flexible, you just bend it round to fit. It may help to do that in sections. It may help to make cuts on the inside surface - some deep, some less deep - perpendicular to the curve. The number of cuts will depend on the thickness of the blade (or the gap it creates).

I don't know about advanced insulating materials, but simple insulation usually has lots of small air pockets, because air is a poor conductor. So provided any unfilled gaps are small, I would not expect them to reduce the effectiveness of the insulation.

I was thinking something like the attached photo, with many pieces shaped as so to fill the whole area, With the material only being 80mm I can make the space between slightly small to 7.5 or so as I will still fall far within my heat loss limit.

 

[Merlin3189]

That looks ok, if pieces cut from your sheet will fit like that.

If you need to calculate the size of the pieces, you know the outer surface is the circumference of a 18.4cm circle and the inside a 10.4cm circle.
So about 65.3 cm inside and 116cm outside.
If you decide to have 100 slices then each is 1.1cm one side and 0.6cm on the other. The angle of cut is 1.8^o so that the sides form a 3.6^o angle, 1/100 of a circle.
The more pieces you have, the easier they will fit, but it may be difficult to cut them accurately and you could alter their insulating property.

But these are rough calculations based on geometry. Depending on the material, the tools and processes used to cut them, you may need to vary them a bit.
For example, if you need to chamfer the outer corners to make the pieces fit, you may want to make them slightly wider at the top. But the difference is probably less than the accuracy with which you can cut many materials.

 

[Jt2015]

Would you mind showing the calculations so I can have a play with the figures? and the fit is not to important as in the past we have just cut wedges and forced them to fit, This time I was just trying to go for a more organised approach, The less slices the better

 

[Merlin3189]

Outer radius = R out = 18.4 cm Inner radius = R in = 10.4 cm Thickness = Th = 8cm
Number of pieces = N
Half angle = HA = 180^o / N
Top width = TW = 2 x Rout x tan(HA)
Bot width = BW = 2 x (Rout - Th) x tan(HA)
Top chamfer = Rout /cos(HA) - Rout
Bot chamfer = Rin - Rout + Th
Bot gap = (Rout - Th)/cos(HA) - Rin

But seeing your comment about "cut wedges and force them to fit" , I wonder whether it is worth trying to be accurate?
If there is enough "give" in the material, just cut say 30 rectangular strips about 4cm. If the last one is a bit too tight, trim that one.

And I'd still consider just getting a can of polyurethane foam.

 

[Jt2015]

View attachment 89067
Outer radius = R out = 18.4 cm Inner radius = R in = 10.4 cm Thickness = Th = 8cm
Number of pieces = N
Half angle = HA = 180^o / N
Top width = TW = 2 x Rout x tan(HA)
Bot width = BW = 2 x (Rout - Th) x tan(HA)
Top chamfer = Rout /cos(HA) - Rout
Bot chamfer = Rin - Rout + Th
Bot gap = (Rout - Th)/cos(HA) - Rin

But seeing your comment about "cut wedges and force them to fit" , I wonder whether it is worth trying to be accurate?
If there is enough "give" in the material, just cut say 30 rectangular strips about 4cm. If the last one is a bit too tight, trim that one.

And I'd still consider just getting a can of polyurethane foam.

Haha, I'm considering the foam option myself but due to the application of the the insulator it needs to operate at high temperature and be very water resistant.
One question with the bottom chamfer calculation, EDIT :Never mind, my mistake.