Sizing sheet material for a solar conical concentrator

by Stephen Hewitt | Published 31 August 2025 | Last updated 27 February 2026

An earlier article contained ideas for a solar cooker with a conical reflector [CONE]. This article continues by deriving the sizes of sheet material needed to make a given size of conical reflector. For simplicity it is restricted to the proposed cone angle of 45°, but it should be easy to generalise to any angle.

It assumes that the reflector will be made of (four or so) equal size segments or sectors. The first part of this article finds the dimensions of the sheet material needed if each segment is cut from a separate rectangular sheet (Figure 1). The second part finds how much sheet material can be saved by instead cutting two segments from a single, larger rectangular sheet. This allows the second segment to overlap the rectangle used for the first segment and utilise some material that would have been an off-cut from the first segment (Figure 2).

The shape on the flat sheet

The shape that is cut out from the flat sheet to make a conical reflector is a sector of the area between two concentric circles. To understand why, consider rolling a piece of paper to make a cone, where the paper is rolled so tightly at the apex of the cone, let's call it O, that there is only one single point of paper there. Then every point on the rim of the cone is an equal distance along the paper from this point, so when you unroll the paper flat the rim must be on an arc of a circle centred at O. The same argument applies to the base of the reflector.

Variables used

α

This is defined here as half the angle subtended by one segment of the reflector on the flat sheet, shown in Figure 3. For the 45° cone considered here α = 360°/(2n√2)

n

The number of separate segments or sectors that the reflector will consist of

O

In the flat plane of the sheet material, this point is the centre of two arcs, one defining the edge of the reflector at the rim and one the edge at the base

r

The radius of the rim of the reflector in its finished, 3-dimensional shape

b

The radius of the base of the reflector in its finished, 3-dimensional shape

Dimensions of a flat sheet for one segment

Now in the case of the cone of angle 45° considered here, with radius at the rim of r we can see by Pythagoras on a cross section along the axis of the cone that the distance from O to the rim is r√2 and that the distance from O to the base is b√2. (Figure 3 of [CONE] shows a cross-section along the axis.)

The circumference of the rim is 2πr, and the angle subtended by the entire rim at point O on the flat sheet must correspond to this length but with (larger) radius r√2, meaning it will be

angle = 2πr/(r√2) = π√2 radians

In other words this angle is 360°/√2, which is about 254.6°. This angle will be divided into n equal segments, so if for example n = 4, then each segment subtends an angle of about 63.6°. It is convenient to define the angle α as half this angle, shown in Figure 3. For a 45° cone,

α = 360°/(2n√2)

This means for a 45° degree cone with 4 segments, α is about 31.8°. Note it depends only on the cone angle, here always 45°, and the number of segments you choose to make the reflector with.

One of these segments is shown in Figure 1. A corresponding half segment is in Figure 3. The following will derive simple expressions for dimensions of sheet material needed for a single segment, given b, r and n.

Length

From Figure 3 length of a sheet for one segment = 2PQ and

PQ = AW = OA.sin(α)

But, as explained above

OA = r√2

So substituting this gives:

PQ = (r√2)sin(α)

The length of the sheet for a full segment is twice this, so

The length of a sheet for one segment = 2sin(α)r√2

Width

From Figure 3, the width of the sheet is QT = OQ - OT.

Now OT = OD.cos(α) and, as already noted OD is the radius of the arc for the base, b√2. Substituting these gives the following.

OT = b.cos(α)√2

The following, similar expression is not needed here but will be used in the next section:

DT = b.sin(α)√2

Also, the radius of the arc for the rim OQ = r√2 as also noted. Substituting these gives the following.

Width of the sheet for one segment = r√2 - bcos(α)√2

Dimensions of a flat sheet for two segments

Figure 4 shows two segments marked out on the same sheet in a way that tries to minimise usage of the material. Clearly the length PQ is unaltered by this arrangement. The saving is the distance VT which is a reduction in total width.

By Pythagoras,

O'D² = O'T² + DT²

But, as noted above,

DT = b.sin(α)√2

And O'V = O'D = r√2, being the radius on the flat surface for the rim.

Substituting for O'D and DT,

O'T² = 2r² - 2(bsin(α))²

O'T = √(2r² - 2(bsin(α))²)

Now from Figure 4,

VT = O'V - O'T

VT = r√2 - √(2r² - 2(bsin(α))²)

VT = (r - √(r² - (bsin(α))²))√2

This VT is the reduction in width, so the width actually needed by the two segments arranged as in Figure 4 would be twice the width derived in the previous section minus this reduction.

Practical application example

I bought the advertised “semi-circular aluminium garden hoops” intended for cloches discussed in the previous article [CONE]. They are made from tubes of 2cm diameter. Trimmed down to exactly a semi-circle each, two of these can form a circular hoop which will go around the rim of the reflector. The internal diameter of the hoop, where the reflector will be supported, measures 1.53m. The rim radius r is therefore:

r = 0.765m.

For now, assuming that the reflector will be made of four segments, α is about 31.8° as noted before. The expression for length derived above was 2sin(α)r√2, so:

length = 2*sin(31.8°)*0.765*1.414 = ~1.14m

This length does not depend on the height of the reflector but only its outer rim size.

For the width of the material it is necessary to decide the height of the reflector. Because the cone angle is 45°, the height of the reflector is the difference between the radius of the rim and radius of the base.

height = r - b

For now let's assume this height will be about 23cm, which is slightly more than the height of the bottles intended to be used and so the base radius will be (say)

b = 0.535m

The expression for width derived above was r√2 - bcos(α)√2, so:

width = 1.414*(0.765 - 0.535*cos(31.8°)) = ~0.44m

The expression derived above for the potential saving in width from cutting two segments from the same sheet as shown in Figure 4 was (r - √(r² - (bsin(α))²))√2, so:

potential width saving = (0.765 - √(0.765*0.765 - (0.535*sin(31.8°))²))*1.414 = ~0.076m

So how do these calculations affect real purchase decisions?

The reflector itself can be a polypropylene sheet covered in metallised plastic, as in an earlier parabolic trough, or, perhaps, metallised film with no further support.

In 2026 one UK supplier of 1mm polypropylene sheets, trentplastics.co.uk, was offering only sizes up to 1m x 1m. The length calculated above is 14cm more than 1m, so it cannot be squeezed down to fit a 1m sheet. To use those, the design would have to change to five segments rather than four.

However there were other suppliers that offer up to 1.2m x 1.2m such as simplyplastics.com and edplastics.co.uk.

With the 4 segment design and the figures calculated, one 1.2m x 1.2m sheet could provide 3 of the segments. That is because the potential width saving from packing three segments together as two of them are packed together in Figure 2 yields two savings, reducing the total width by 2 x 0.076m or about 0.15m. The naive way of cutting them would require 0.44m width for each one, or a total of 1.32m but with this reduction it falls to 1.17m, which fits on a 1.2m sheet.

However for a one-off production this still leaves the 4th segment, which might need a big sheet all to itself. The idea that a segment might be too small and so waste unused parts of a sheet also suggests the possibility of making a cone with segments of unequal angle. Then the calculations could be made backwards, starting with the standard sheet size and calculating the segment angles to fill it. For a cone with n segments, you would make (n - 1) of the segments of equal angle. You maximise this angle to use as much of a standard sized sheet as possible. Then you would calculate what angle remains for the final segment to complete the cone, and whether it was now small enough to fit onto some smaller sheet, that perhaps equal-sized segments would not have fitted. Further work is needed to investigate this. It depends also on whether the construction of the cone allows unequal segments.

In this case it does not necessarily arise because if 3 segments are taken from a 1.2m x 1.2m sheet, this is already a tight fit.

For the metallised film, various kinds are available on rolls of 1.2m width from 1-hydroponics.co.uk, so this turns out to be a good fit for the length of 1.17m calculated here. The segments will go lengthwise across the width of the roll and savings can be made by packing the segments together along the roll in the manner of Figure 2.

References

[CONE]

Ideas for a solar cooker with a conical concentrator, Stephen Hewitt, 17 June 2024