DIY maths/trig

[Jethro_uk]

Sitting in our smoking (barbecue) shelter last night, I pondered on building a conservatory with the same dimensions, but a bay at one end.

It then dawned on me that unless I am missing something basic, there's a hairy bit of maths to do.

Starting with the intention that the bay be constructed with 3 equal width panels.

You start with a semi circle, whose radius will be equal to the width of the shelter (c. 1.5m) and that extends out. You then have to place within that semi circle a trapeziod whose base is c. 1.5m and the other 3 sides are of equal length.

Having seen how medieval castle builders used string and chalk to mark out complex patterns, I suspect this would be the simplest route.

Or, to rejig the question. How far from the base of the bay will the middle panel be (clue, it's *not* the radius of the semi circle).

[Jethro_uk]

Googling reveals

angles.html

however, that's the sausage machine method. How did they get to that process ?

[News]

In message , Jethro_uk writes

Not terribly accurately, apparently.

1 Measure the width of the central pane. For instance, this pane may measure 25 inches. 2 Measure the entire bay's width. For instance, the bay may be 50 inches in width. 3 Subtract the central pane's width from the whole bay's width. With this example: 50 - 26 = 24 inches.

Perhaps I am missing something, but it seems that 25 inches suddenly becomes 26 inches.

[Davey]

At least the width of the bay stayed the same.

[The Natural Philosopher]

well its r sin 60 or summat innit.

sinve a bay of equal panels will be a trisection of the half circle, each panel will cover and be a chord to a 60 degree arc.

[The Natural Philosopher]

further to that the panels will each be half the width of the panel span, since 60 degrees is an equilateral triangle

So for a 1.5m base you need three 0.75m panels

[Mike Barnes]

I assume you want the vertices *on* the semicircle.

If that's so, the angle subtended by each pane is 60 degrees, and the ratio of the projection to the radius is the sine of 60 degrees, which Google tells me is about 0.866.

So for 1500 mm width, the answer is 1500/2*0.866 = 649 mm. Call it 650.

E&OE

[Brian Reay]

You need to draw a diagram.

First a trapezium, short parallel side 50 from the example they use.

The long side is 50+24, so mark the 50 in the centre section and two bit

12 at each end. 'Dot in' vertical lines to divide the trapezium into a rectangle and two, right angled, triangles.

Mark the dotted line 14.

Now, form another rectangle, at one end, using the right angle triangle as your guide and the 'sloping' side of the trapezium as its diagonal.

The angle they have found is between the diagonal and the side of your last rectangle.

Then that is equal due to 'alternate angle theorem' to part of the angle you require. The other part is a right angle so you add 90

The method in the example isn't as short as it could be, there are a few steps which you could avoid, but that is how the they've done it.

[Dave Plowman (News)]

Get something like DraftSight 2D CAD prog (it's free).

Draw it out to scale using the measurements you know. Easy then to get all the required dimensions and angles.

[Jethro_uk]

Correct

Differs from TNPs 0.75m ...

[Jethro_uk]

When thinking about it last night, I remembered the word "chord" from years ago ...

[Jethro_uk]

Not Mike Barnes 650 ?

[The Natural Philosopher]

yeah, that's the way for brainless idiots who cant do sums to do it.

[The Natural Philosopher]

No it doesnt.

.75m is the panelsize, the final DEPTH of the bay is r sin 60 which is what he calcualates above.

[Robin]

FWIW I'm with TNP. Each of the 3 panes is the side of a regular hexagon. Them's 6 equilateral triangles in there. So the diametetr which is your "base" is 2 sides. You say that's 1.5m So each panel is

0.75m.

Much more fun where the base is not a diameter of the circle but an arbitrary chord. But you'd need to pay me to try to excavate and renovate the A-level trig - and then wait a while or 3 :(

[Robin]

Sorry! I should have read on.

I think TNP and I answered the wrong question. You want the distance from the base to the *middle of a panel*? Then that is, I agree,

649.5mm. (Think isosceles triangle with sides 0.75, 0.75/2 and the distance D to the middle of the panel. Then =SQRT(0.75^2-(0.75/2)^2) )

[The Natural Philosopher]

No, that is the DEPTH of the bay there. How far from the base of the bay will the middle panel be = 0.75 sin(60)= about = 650mm

but the panels are 0.75m long. for an infinitely thin panel ;-)

There are of course many other equal sized panels between 0.5 and 1.5 meters that will make a trapezoidal bay, but only one with nicely balanced angles.

a 45 degree one would be a bit shallower.

The original question starts with an assumption that may in fact not be valid or necessary

"You start with a semi circle"

There is no need to start with a semi-circle at all, and other equal length trapezoids will still fit the 1.5m base, but give different depths and different panel sizes.

It is the requirement to be chords of a *semi* circle that defines the single 60 degree solution, if the panel lengths are equal

Likewise if it is a semi circle and the panel lengths are undefined, then there are an infinite number or solutions as well from no bay at all to a sort of pointed abutment with no parallel wall at all and all types in between,.

Only the semi circle AND the equal panel lengths case has one single solution. Well two if you count the inverted case where the bay is built into rather than out side the existing house ;-)

,

[The Natural Philosopher]

There is no disagreement: we are calculating different things.

panel size versus depth.

[Mike Barnes]

Agreed, three 750 mm panels.

But the question was how far the bay would project, and that's the question I answered: 650 mm.

But that's just an estimate. I'd add on a few mm for extras related to the thickness of the panels, compared to the idealised lines of the geometric model.

[Jethro_uk]

Ah, of course. If I'd envisage a *hexagon* it would have made it much easier.