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?



--
Dave - The Medway Handyman www.medwayhandyman.co.uk

"Tim Streater" wrote in message
...
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.

or even a calculator

cos (75) is 0.258

so rounding it down a bit it's

Y +0.25(Z)

tim

On Sat, 27 Jul 2013 11:07:51 +0100, 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?

SOHCAHTOA.

--
Frank

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

On 27/07/2013 11:07, 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?




Tan @ 15 degrees is your calculation = 0.268. That is then the ratio
between the width of the awning and the increase in height.

The Medway Handyman :
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?

The easy way to answer this and similar questions is to make a scale
drawing. It doesn't have to be anything fancy, all you need is pencil,
paper, and ruler.

A spreadsheet would be quicker and more accurate but a drawing would be
accurate enough and would give you a better feel for what the result
would look like.

--
Mike Barnes

On Sat, 27 Jul 2013 11:07:51 +0100, 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?

I assume you mean 15 degrees below horizontal.
"Depenfing on the width" - of what, measured how?
I assume you measure the extension E down the sloping face, not
horizontally.

X=Y+Esin15
i.e. X=Y+(0.26 x E)
Your example would be X=2.1 + 0.26x2, i.e. 2.62m

If you can add the Y in your head, all you need to remember is that X
is 0.26xE above Y.
--
Dave W

On Sat, 27 Jul 2013 12:17:39 +0100, Frank Erskine
wrote:

SOHCAHTOA.


Oswald Has Been Here Obtaining Beer (but nothing to remember which was
which - Sine, Cosine, Tan) B=base
[from at least 50yrs ago ]

--
AnthonyL

On 27/07/2013 14:15, Tim Streater wrote:
In article ,
Ramsman wrote:

On 27/07/2013 11:54, Tim Streater wrote:

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.

Well, only Our Dave knows whether the Z refers to the length of material
that has unrolled from the sunroof (my interpretation) or refers to the
horizontal distance extended (your interpretation).

Either way, once you have the formula, no spreadsheet is needed. I just
typed my formula straight into Google and it gave me the answer.


As we were always told at school, read the question. TMH said 'I want to
make up a spreadsheet to calculate this for any height "Y" and any
extension.' Hence my reply and my formula.

--
Peter

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?



Personally I'd do a scale drawing and use that. This sort of thing
http://www.wrightsaerials.tv/referen...ed-by-roof.pdf

Bill

On 27/07/2013 15:49, Tim Streater wrote:
In article ,
Ramsman wrote:

On 27/07/2013 14:15, Tim Streater wrote:

Either way, once you have the formula, no spreadsheet is needed. I just
typed my formula straight into Google and it gave me the answer.

As we were always told at school, read the question. TMH said 'I want
to make up a spreadsheet to calculate this for any height "Y" and any
extension.' Hence my reply and my formula.

And I'm pointing out that you don't need one. For any given extension,
just get google to do it or use your handy pocket calculator. A
spreadsheet is overkill - unless you are learning spreadsheets. But then
we weren't told that.


'I want to make up a spreadsheet to calculate this for any height "Y"
and any extension.'

Perhaps you don't need one, but that's what he clearly asked for, so
that's what I gave him. Having set it up, you only have to enter the
values and recalculate each time.

--
Peter

On 27/07/2013 11:07, 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?





You can make up an Excel spreadsheet as follows:

Put in awning front height - in metres - in column (say) D3

Then put the awning width (by that I mean front to back) - again in
metres - in column D5

Then put the slope - in degrees - in column D7

Your final column, which will give the height of the rear of the awning,
will need the following formula: =TAN(D7*PI()/180)*D5+(D3)

On 27/07/2013 14:15, Tim Streater wrote:

Well, only Our Dave knows whether the Z refers to the length of material
that has unrolled from the sunroof (my interpretation) or refers to the
horizontal distance extended (your interpretation).

Either way, once you have the formula, no spreadsheet is needed. I just
typed my formula straight into Google and it gave me the answer.

If the extension is the horizontal reach, then it might be simpler to
simply think of the slope as a rise in terms of "1 in 4" or similar...

So a 2m extension translates as 0.5m rise at that going.


--
Cheers,

John.

/================================================== ===============\
| Internode Ltd - http://www.internode.co.uk |
|-----------------------------------------------------------------|
| John Rumm - john(at)internode(dot)co(dot)uk |
\================================================= ================/

In article , Huge
scribeth thus
On 2013-07-27, 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?

Everything you need to know is in this article;

http://en.wikipedia.org/wiki/Right_Angle_Triangle



Yes fine but often it seems that Wikipedia goes into some complex maths
rather too soon in the explain process. I 'm sure 'tho I can't remember
it there is a very good site around that has a few worked out examples
and shows you how those are done thats rather more educative than this
one...

--
Tony Sayer

On 27/07/2013 11:07, 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?



This is the easy way (although the calculation isn't all that complicated)
http://www.cleavebooks.co.uk/scol/calrtri.htm

in 1243759 20130727 110751 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?

If you've got an Android phone then I recommend this :

https://play.google.com/store/apps/d...trianglesolver

On Sat, 27 Jul 2013 12:57:02 +0100, Dave W
wrote:

I assume you mean 15 degrees below horizontal.
"Depenfing on the width" - of what, measured how?
I assume you measure the extension E down the sloping face, not
horizontally.

X=Y+Esin15
i.e. X=Y+(0.26 x E)
Your example would be X=2.1 + 0.26x2, i.e. 2.62m

If you can add the Y in your head, all you need to remember is that X
is 0.26xE above Y.

Even easier, as sin15 is pretty much 1/4 (and how accurate is your 15
anyway), X is E/4 higher than Y.
--
Dave W

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?

I am amazed that Unbeliever has not yet called you a **** for not knowing
such things.

--
Adam

On Sunday, 28 July 2013 15:32:07 UTC+1, wrote:
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?



I am amazed that Unbeliever has not yet called you a **** for not knowing

such things.



--

Adam

Download a free copy of 'Sketchup' and draw it in that. It's the easiest draughting programme ever. You can check all the dimensions you like direct on the model

On 28/07/2013 15:32, ARW wrote:
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?

I am amazed that Unbeliever has not yet called you a **** for not knowing
such things.

Can't risk that when so far out of his depth himself...


--
Cheers,

John.

/================================================== ===============\
| Internode Ltd - http://www.internode.co.uk |
|-----------------------------------------------------------------|
| John Rumm - john(at)internode(dot)co(dot)uk |
\================================================= ================/

On 28/07/2013 15:32, ARW wrote:
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?

I am amazed that Unbeliever has not yet called you a **** for not knowing
such things.

I'm amazed he hasn't called you a **** for knowing me.

Perhaps he has pegged it at last.

--
Dave - The Medway Handyman www.medwayhandyman.co.uk

On 27/07/2013 14:15, Tim Streater wrote:
In article ,
Ramsman wrote:

On 27/07/2013 11:54, Tim Streater wrote:

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.

Well, only Our Dave knows whether the Z refers to the length of material
that has unrolled from the sunroof (my interpretation) or refers to the
horizontal distance extended (your interpretation).

Either way, once you have the formula, no spreadsheet is needed. I just
typed my formula straight into Google and it gave me the answer.

Sorry, "Z" would be the fabric/hypotenuse.

The 1 to 4 m would be the second longest side of the triangle, the
third/shortest side of the triangle would be the height difference.

Reason I was going to do a spreadsheet is so I could have a handy paper
chart for site use. No Google involved.

--
Dave - The Medway Handyman www.medwayhandyman.co.uk

On 28/07/2013 15:27, Dave W wrote:
On Sat, 27 Jul 2013 12:57:02 +0100, Dave W
wrote:

I assume you mean 15 degrees below horizontal.
"Depenfing on the width" - of what, measured how?
I assume you measure the extension E down the sloping face, not
horizontally.

No, horizontally.

X=Y+Esin15
i.e. X=Y+(0.26 x E)
Your example would be X=2.1 + 0.26x2, i.e. 2.62m

If you can add the Y in your head, all you need to remember is that X
is 0.26xE above Y.

Even easier, as sin15 is pretty much 1/4 (and how accurate is your 15
anyway), X is E/4 higher than Y.

I haven't explained this clearly enough. My bad.

The awning is bolted to the wall & the front edge when extended, is
lower. The canvas is at a 15 degree slope. In some cases the front of
the awning has to be at a certain height to miss an obstacle.




--
Dave - The Medway Handyman www.medwayhandyman.co.uk

On 27/07/2013 17:15, Farmer Giles wrote:
On 27/07/2013 11:07, 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?





You can make up an Excel spreadsheet as follows:

Put in awning front height - in metres - in column (say) D3

Then put the awning width (by that I mean front to back) - again in
metres - in column D5

Then put the slope - in degrees - in column D7

Your final column, which will give the height of the rear of the awning,
will need the following formula: =TAN(D7*PI()/180)*D5+(D3)

Thanks - that seems to work!


--
Dave - The Medway Handyman www.medwayhandyman.co.uk

On Mon, 29 Jul 2013 19:07:35 +0100, The Medway Handyman
wrote:

On 28/07/2013 15:27, Dave W wrote:
On Sat, 27 Jul 2013 12:57:02 +0100, Dave W
wrote:

I assume you mean 15 degrees below horizontal.
"Depenfing on the width" - of what, measured how?
I assume you measure the extension E down the sloping face, not
horizontally.

No, horizontally.

X=Y+Esin15
i.e. X=Y+(0.26 x E)
Your example would be X=2.1 + 0.26x2, i.e. 2.62m

If you can add the Y in your head, all you need to remember is that X
is 0.26xE above Y.

Even easier, as sin15 is pretty much 1/4 (and how accurate is your 15
anyway), X is E/4 higher than Y.

I haven't explained this clearly enough. My bad.

The awning is bolted to the wall & the front edge when extended, is
lower. The canvas is at a 15 degree slope. In some cases the front of
the awning has to be at a certain height to miss an obstacle.

As you liked Farmer Giles's solution, I now change my first solution
to X = 0.27E above Y, where E now represents the horizontal distance
of the front from the wall.
--
Dave W

On 29/07/2013 19:12, The Medway Handyman wrote:
On 27/07/2013 17:15, Farmer Giles wrote:
On 27/07/2013 11:07, 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?





You can make up an Excel spreadsheet as follows:

Put in awning front height - in metres - in column (say) D3

Then put the awning width (by that I mean front to back) - again in
metres - in column D5

Then put the slope - in degrees - in column D7

Your final column, which will give the height of the rear of the awning,
will need the following formula: =TAN(D7*PI()/180)*D5+(D3)

Thanks - that seems to work!




Glad to help.

I'd be grateful if you could now do me a favour. Please call it 'maths',
and not the irritating Americanism 'math'.

On 30/07/2013 16:13, Farmer Giles wrote:
On 29/07/2013 19:12, The Medway Handyman wrote:
On 27/07/2013 17:15, Farmer Giles wrote:
On 27/07/2013 11:07, 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?





You can make up an Excel spreadsheet as follows:

Put in awning front height - in metres - in column (say) D3

Then put the awning width (by that I mean front to back) - again in
metres - in column D5

Then put the slope - in degrees - in column D7

Your final column, which will give the height of the rear of the awning,
will need the following formula: =TAN(D7*PI()/180)*D5+(D3)

Thanks - that seems to work!




Glad to help.

I'd be grateful if you could now do me a favour. Please call it 'maths',
and not the irritating Americanism 'math'.

I stand corrected :-)


--
Dave - The Medway Handyman www.medwayhandyman.co.uk

The Medway Handyman wrote:
On 28/07/2013 15:27, Dave W wrote:
On Sat, 27 Jul 2013 12:57:02 +0100, Dave W
wrote:

I assume you mean 15 degrees below horizontal.
"Depenfing on the width" - of what, measured how?
I assume you measure the extension E down the sloping face, not
horizontally.

No, horizontally.

X=Y+Esin15
i.e. X=Y+(0.26 x E)
Your example would be X=2.1 + 0.26x2, i.e. 2.62m

If you can add the Y in your head, all you need to remember is that
X is 0.26xE above Y.

Even easier, as sin15 is pretty much 1/4 (and how accurate is your 15
anyway), X is E/4 higher than Y.

I haven't explained this clearly enough. My bad.

The awning is bolted to the wall & the front edge when extended, is
lower. The canvas is at a 15 degree slope. In some cases the front of
the awning has to be at a certain height to miss an obstacle.

When I helped fit the awning at my parents house I let my Dad "borrow" the
SDS drill that I bought him for Christmas:-)

--
Adam

On 30/07/2013 16:13, Farmer Giles wrote:

I'd be grateful if you could now do me a favour. Please call it 'maths',
and not the irritating Americanism 'math'.

cough This is arithmetic.

Andy

In article , Vir Campestris
writes
On 30/07/2013 16:13, Farmer Giles wrote:

I'd be grateful if you could now do me a favour. Please call it 'maths',
and not the irritating Americanism 'math'.

cough This is arithmetic.

Not trigonometry then B-)
--
fred
it's a ba-na-na . . . .

On 30/07/2013 22:03, Vir Campestris wrote:
On 30/07/2013 16:13, Farmer Giles wrote:

I'd be grateful if you could now do me a favour. Please call it 'maths',
and not the irritating Americanism 'math'.

cough This is arithmetic.

Andy

cough Trigonometry is maths. The final calculation is arguably
arithmetic, but the process that defined the calculation is definitely
mathematics.