On 27/07/2013 11:54, Tim Streater wrote:
In article ,
The Medway Handyman wrote:

An awning is attached to a wall at a height "X".


The front edge is lower because the awning slopes by 15 degrees.


Depending on the width, the awning could extend by anything from 1m to
4m.

If I want the front edge at a specific height "Y", what would distance
"X" be?

I want to make up a spreadsheet to calculate this for any height "Y"
and any extension. So, if the client wants the front of the awning
2.1m off the ground and it extends 2m, how high would "X" be?

Well if it slopes 15deg then the angle it makes with the wall is 90-15 =
75deg. So if it extends out Z (which you say is 1m to 4m), then that's
the hypotenuse of the triangle in which case the vertical height of the
triangle is given by:

H / Z = cos 75

So:

H = Z * cos 75

But you said the front edge is up by 2.1m so that gives you:

X = 2.1 + H

or X = 2.1 + (Z * cos 75)

if Z=2 then I type this into Google:

2.1 + (2 * cos (75 deg))

and it gives me 2.617....


No need for a spreadsheet at all.

Here's the diagram

.
|\
| \
| \
| \ Z
| \
|H \
| \
| \
| \
.---------.
|
|
|
|
| 2.1
|
|
|


My understanding is that 2m is the horizontal distance ('it extends out
Z'), so it's different from that.

|\
| \
| \
| \
| \
|X-Y \
| \
X| 15\
---------
| Z
|
|
| Y
|
|
|

(X-Y)/Z = tan 15
..
..
..
X = Y + Z (tan 15)

Spreadsheet with Y in A1, Z in B1, then C1 is =A1+(B1*TAN(RADIANS(15)))
which gives 2.64 for the given values.

HTH
--
Peter