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June 4th, 2018, 03:17 AM   #1
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Converting angle in inches to degrees

Hi forum, my window air conditioner instructions indicate that the unit must be tilted back 1/4" for proper drainage. I'd like to figure this out in degrees, since I took a college trig a few years ago and feel like I should know how to solve something so basic. My initial attempt has me drawing a unit circle with with a right angle measuring .25 on the y axis.

How to approach this problem? Thanks in advance.
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June 4th, 2018, 05:29 AM   #2
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The instructions are ambiguous, as they do not state which part of the air conditioner has to move 1/4” when the air conditioner is tilted.
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June 4th, 2018, 06:23 AM   #3
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Hi, a window air conditioner is one, self-contained unit that tilts back so that excess condensate drains outside. The tilt of the unit is specified as 1/4" angle downward (to the outside).

If someone might help me get started converting that 1/4" tilt to a degree angle I'd be most appreciative.
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June 4th, 2018, 07:28 AM   #4
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Quote:
Originally Posted by paulm View Post
Hi, a window air conditioner is one, self-contained unit that tilts back so that excess condensate drains outside. The tilt of the unit is specified as 1/4" angle downward (to the outside).

If someone might help me get started converting that 1/4" tilt to a degree angle I'd be most appreciative.

Assuming the air conditioner is hinging on the bottom mating corner, we'd need to know the length of the unit to calculate the angle. However measuring total inch drop at the back of the unit is more accurate in all practicality.
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June 4th, 2018, 07:33 AM   #5
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What is the distance from the front to the rear of the unit along the bottom?
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June 4th, 2018, 09:13 AM   #6
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What is the distance from the front to the rear of the unit along the bottom?
Thanks, that got me started. The back-half of unit is already angled, and instructions indicate measuring from where window meets the angled part. So the angled back-half of unit is 7.8". I found the last (adjacent) side of triangle through Pythag. Ther. (1/4)^2 + x^2 = (7.8 )^2, thus adjacent side = 7.79, and then solving cos = .998 and acos = 3.62°, which is exactly the angle of back-half of unit, using bubble level.

Basically the unit could be level (but not at all tilting forward) and drain properly, given the already-angled back.

I wanted to know how to approach this using trig so thanks for reminding me how to get started.

Last edited by paulm; June 4th, 2018 at 09:16 AM.
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