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I'm looking to build a small covered patio area on our property, next to our pool. It'll be open on all sides, and covered by a simple skillion/single-sloped roof. So something like in this photo:

Example

The idea behind the single sloped roof is that we'd like to be able to put a bunch of PV panels up there to generate some solar energy.

How can I combine the roof dimensions, with the angle/pitch of the roof, to determine the amount of area that will actually be covered? Suppose the roof is 20 feet long by, say, 15 feet wide. If it were flat that would be easy. But if the roof is angled at say, 30 degrees, then that will change the actual covered space below it. How would I go about calculating that exactly?

Crossposted to Mathematics

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    Rather than specifying the roof area and then calculating the covered space, wouldn't you do it the other way around? – Daniel Griscom Mar 27 '16 at 23:35
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    Trigonometry? – Comintern Mar 27 '16 at 23:40
  • This is more of a basic algebra question than a home improvement question. Lookup Pythagorean Theorem. – Edwin Mar 27 '16 at 23:40
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    Mazura, yes, that's exactly it! Because of the PV panels, the roof size and angle will be what determines the covered area underneath, not the other way around. But yes, I'll just go ask over at Math SE, that might be a better place to try. – asjd Mar 28 '16 at 0:50
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    @asjd Please don't post the same question to multiple StackExchange sites. If you think the question would fit better somewhere else, please flag it for moderator attention, and we'll migrate it. Thanks. – Tester101 Mar 28 '16 at 13:33
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area under roof = L x W x cos 30

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    I would add that since it doesn't sound like OP has picked a specific angle, for angles other than 30º it would be cos(ANGLE). – Hank Mar 28 '16 at 15:10
  • this may sound dopey, but when you guys refer to "op" what does that mean? original poster? – personal privacy advocate Mar 28 '16 at 15:42
  • @personalprivacyadvocate Yes, OP = Original Poster. – Tester101 Mar 28 '16 at 16:10
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    ...and make sure the calculator is in degrees mode (30 radians will give a very different answer.) – Ecnerwal Mar 29 '16 at 2:43
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If you built the roof 20' long by 15' wide, then tilted the roof at a 30° angle. The roof would cover about 260 sq. ft., or a patio 20' long by about 13 ' wide.

Law of sines for 15' roof

If you want to have a 20'x15' patio, and cover it with a 30° roof. Then the roof will have to be about 17' 4" wide, with an area of 346.41 sq.ft.

Law of sines 15' patio


To appease Henry Jackson, you could also do the calculations like this.

Cosine

  • This is true but you have really taken the "long way around". cos(A) = b / c is basically the definition of cos and give the answer immediately. – Hank Mar 28 '16 at 15:15
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    @HenryJackson Edited just for you. – Tester101 Mar 28 '16 at 16:08
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Building backwards isn't a good plan & you won't ever be able to get materials to "exact" in the real world. But, your 300-s.f. roof covers a 20' x 12'-11 & 11/64" patio/room (to be a hair shy). Multiply those dimensions with the 1.16 pitch (from picture below) to arrive back at your 300s.f.

Here's a Decimal Converter that translates the pitch dimension (12'-11 & 11/64") down to 12.930. The 12.930 is affirmed by then dividing your 15 by the 1.16 multiplier to actually end up with 12.931 (now, to be a hair heavy). 100ths or 1000ths would need to be pointlessly instituted to get the multiplying & dividing numbers to specifically match.

Below is the Degree Of Angle equivalency (your 30° would be closest to a 7/12 pitch) & their Multipliers to your Length & Width tight-to-building box measurements. You'd have to further figure in any roof overhang reductions.

roof pitch

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