Finding the Angle of Solar Collectors on a Sloped RoofDate: 08/30/2008 at 03:13:19 From: Neill Subject: Angle of solar collectors on a sloped roof After three days of thinking about it, making paper models and drawing sketches, I am absolutely stumped. On the roof of my house I have solar collectors and would like to know which way do they face (which way does the fall line that rain would follow face)? What is the angle of the fall line to the horizon? The collectors are shown at http://www.hogarth.de/heating/about.php. The roof has an angle of 26 degrees to horizontal and the collectors have an angle to the roof of 43 degrees. The collectors are at right angles to the roof. The ridge of the roof runs due south/north. My guesses are about 26 degrees from south and about 48 degrees to the horizon. Ideally I would like to know the answers and how to work this out so that I can do it myself in future for other collectors. From school I remember vectors and am sure I must be able to use them to solve this problem some how. Greetings from Bavaria! Thank you! Neill Date: 08/31/2008 at 21:06:41 From: Doctor Rick Subject: Re: Angle of solar collectors on a sloped roof Hi, Neill. You're right, vectors will help--along with rotation matrices: Rotations in Three Dimensions http://mathforum.org/library/drmath/view/51994.html Let's set up coordinate axes: the positive x axis to the east, positive y axis to the north, and positive z axis pointing up. If I understand correctly, the peak of the roof is aligned north-south, that is, parallel to the y axis. Now, here's what we can do. We'll start with the solar collectors horizontal, and define a vector perpendicular to that surface (this is the usual way we identify the orientation of a plane). This "normal vector" is vertical, with components (0, 0, 1). First we tilt the solar panel by the angle beta, which will be the angle the collector makes with the roof. (So far we have a flat roof.) This requires a rotation of beta degrees counterclockwise about the (positive) x axis: / 1 0 0 \ A = | 0 cos(beta) -sin(beta) | \ 0 sin(beta) cos(beta) / Then we rotate the roof (with the solar collector on it) by an angle alpha clockwise about the y axis; this will be considered a negative angle, -alpha. The rotation matrix for this part of the transformation is / cos(-alpha) 0 sin(-alpha) \ B = | 0 1 0 | \ -sin(-alpha) 0 cos(-alpha) / / cos(alpha) 0 -sin(alpha) \ = | 0 1 0 | \ sin(alpha) 0 cos(alpha) / Now we put it all together: we start with the normal vector N (represented as a column vector), multiply it (on the left) by matrix A, then multiply (on the left again) by matrix B. The new normal vector is N' = BAN / -sin(alpha)*cos(beta) \ = | -sin(beta) | \ cos(alpha)*cos(beta) / Finally, we want to know the direction this vector is pointed. The angle between this vector and the vertical (which is equal to the angle between the rain fall line and the horizontal), which I'll call theta, is found by taking the dot product with the vertical vector (0, 0, 1), and taking the inverse cosine (because both the vectors whose dot product we're taking have length 1): theta = arccos(cos(alpha)*cos(beta)) The angle phi in the horizontal plane can be found by projecting N' into the x-y plane (that is, setting the z component to 0), then using the atan2 function found in programming languages and spreadsheets (with some difference in usage, namely the order of arguments) to convert Cartesian to polar coordinates: phi = atan2(y, x) = atan2(-sin(beta), -sin(alpha)*cos(beta)) This will be in radians counterclockwise from east; to change it to degrees clockwise from north, we can use phi = 90 - 180/pi*atan2(-sin(beta), -sin(alpha)*cos(beta)) It's time to test my work! I'll use a spreadsheet (which will require that I switch the arguments to atan2). Using your values alpha = 26 deg beta = 43 deg I get values of theta = 48.90 deg phi = 205.18 deg You estimated that theta was 48 degrees. For phi, your estimate of 26 degrees from south, I presume *west* of south, would correspond to 180+26 = 206 degrees from north. It looks like I may have gotten it right on the first try! Note that I ran through the work quickly (as I was working it out), and did not stop to explain everything. Feel free to ask for more explanation wherever you need it. Hope this helps! - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ Date: 09/01/2008 at 05:00:53 From: Neill Subject: Thank you (Angle of solar collectors on a sloped roof) Thank you! It is nice to have our guesses and models confirmed by you. I will try and fully understand your answer but I think I need to look into a few basics first. Thanks Neill |
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