Ji

"Joe"

31/12/2006 10:23 PM

angles: trig help

I want to double check some wedges I've made for angled cuts. In a 24" long
wedge, how much gain on the square corner is there for each degree of angle?

I know the method is rattling around back there amongst the unused and as
yet unkilled brain cells but I can't seem to come up with it.

Thanks to the more mathmatically inclined.

jc


This topic has 10 replies

Ji

"Joe"

in reply to "Joe" on 31/12/2006 10:23 PM

01/01/2007 3:10 PM

Thanks everyone for the help. In the original post, I neglected to mention
that I was calculating this for the first couple of degrees. I've got it
now.



Have a very happy and healthy New Year.

jc


"Joe" <[email protected]> wrote in message
news:[email protected]...
>I want to double check some wedges I've made for angled cuts. In a 24"
>long wedge, how much gain on the square corner is there for each degree of
>angle?
>
> I know the method is rattling around back there amongst the unused and as
> yet unkilled brain cells but I can't seem to come up with it.
>
> Thanks to the more mathmatically inclined.
>
> jc
>
>

MD

"Morris Dovey"

in reply to "Joe" on 31/12/2006 10:23 PM

31/12/2006 4:38 PM

Joe wrote:
| I want to double check some wedges I've made for angled cuts. In a
| 24" long wedge, how much gain on the square corner is there for
| each degree of angle?
|
| I know the method is rattling around back there amongst the unused
| and as yet unkilled brain cells but I can't seem to come up with it.
|
| Thanks to the more mathmatically inclined.

It's not the same for each degree of angle (the gain going from 0 to 1
degree is a lot larger than the gain when you go from 88 to 89).

Since the tangent of an angle (a) is equal to the rise (y) divided by
the run (x=24), you can calculate:

tan(a) = y / 24

y = 24 * tan(a)

If a1 is your starting angle then

y1 = 24 * tan(a); and
y2 = 24 * tan(a+1)

and the gain for that 1 degree difference is

y2 - y1 = 24 * (tan(a+1) - tan(a))

--
Morris Dovey
DeSoto Solar
DeSoto, Iowa USA
http://www.iedu.com/DeSoto

MD

"Morris Dovey"

in reply to "Joe" on 31/12/2006 10:23 PM

31/12/2006 4:40 PM

Morris Dovey wrote:

| It's not the same for each degree of angle (the gain going from 0
| to 1 degree is a lot larger than the gain when you go from 88 to
| 89).

I think I said that backward. :-(

larger --> smaller

--
Morris Dovey
DeSoto Solar
DeSoto, Iowa USA
http://www.iedu.com/DeSoto

Wi

"Wilson"

in reply to "Joe" on 31/12/2006 10:23 PM

01/01/2007 5:12 AM

OOPS, maybe a little doubt would be good on this one.
G/HYP does = SIN X, but you have two unknowns, since the base is fixed at
24.
G = HYP*SIN X doesn't help, since we don't know the hypotenuse.
As Tin said, TAN X = G/24, so G = 24 TAN X.
I hope I'm not too sleepy to get this right.
WL
<[email protected]> wrote in message
news:[email protected]...
> In article <[email protected]>,
> Joe <[email protected]> wrote:
>>I want to double check some wedges I've made for angled cuts. In a 24"
>>long
>>wedge, how much gain on the square corner is there for each degree of
>>angle?
>>
>>I know the method is rattling around back there amongst the unused and as
>>yet unkilled brain cells but I can't seem to come up with it.
>>
>>Thanks to the more mathmatically inclined.
>>
>>jc
>>
>>
>
> What do you mean by "gain" ? If you mean, for a right triangle with
> angel A between the hypotenuse and a 24" side, that the "gain" is the
> length of the side opposite angle X, then the change in the length of
> that side is not a linear function wrt the angle X. However, it will
> be equal to 24 X sine(A)
>
> --
> Often wrong, never in doubt.
>
> Larry Wasserman - Baltimore, Maryland - [email protected]

Tt

"TinWoodsmn"

in reply to "Joe" on 31/12/2006 10:23 PM

31/12/2006 3:28 PM


"Joe" <[email protected]> wrote in message
news:[email protected]...
>I want to double check some wedges I've made for angled cuts. In a 24"
>long wedge, how much gain on the square corner is there for each degree of
>angle?
>
> I know the method is rattling around back there amongst the unused and as
> yet unkilled brain cells but I can't seem to come up with it.
>
> Thanks to the more mathmatically inclined.
>
> jc
>
>
Assuming the 24" side is included between the right angle and the angle to
be incremented, and the 24" side and the 'gain' side are at right angles to
each other:

Gain in inches = 24*tangent(incremented angle).

For some angles:
1 deg-> .4189516"
2 deg-> .8380985
15 deg -> 6.4307806"
22.5 deg-> 9.5941125"
30 deg -> 13.8564061"
45 deg -> 24.0000000"
89 deg-> 1,374.9590791"
etc.....

Using Windows calculator makes the computation easy once you know the
equation.

Tin Woodsmn

Happy New Year to all of the Wreckers....

Tt

"Tyke"

in reply to "Joe" on 31/12/2006 10:23 PM

31/12/2006 5:52 PM

As Morris Dovey warned in a separate message, the height is not linear for
each degree. Hence you do need to use the formula.

Dave Paine

"Tyke" <[email protected]> wrote in message
news:[email protected]...
>I am not going to attempt ASCII art.
>
> If by the "square corner" you mean the two sides of a right angle triangle
> which are orthogonal (90 degree angle between these sides, then the
> formula is
>
> Height = Length * Tangent(opposite angle).
>
> For Length = 24 inch and opposite angle = 1 deg, then Height = 0.418922
>
> If you have MS Excel, then you can enter a formula
> Cell for Height = =Cell for Length*TAN(RADIANS(Cell for angle))
>
> For some strange reason MS designed Excel to use RADIANS in angle
> functions instead of degrees where 180 deg = PI Radians.
>
> Dave Paine
>
> "Joe" <[email protected]> wrote in message
> news:[email protected]...
>>I want to double check some wedges I've made for angled cuts. In a 24"
>>long wedge, how much gain on the square corner is there for each degree of
>>angle?
>>
>> I know the method is rattling around back there amongst the unused and as
>> yet unkilled brain cells but I can't seem to come up with it.
>>
>> Thanks to the more mathmatically inclined.
>>
>> jc
>>
>>
>
>

Tt

"Tyke"

in reply to "Joe" on 31/12/2006 10:23 PM

31/12/2006 5:50 PM

I am not going to attempt ASCII art.

If by the "square corner" you mean the two sides of a right angle triangle
which are orthogonal (90 degree angle between these sides, then the formula
is

Height = Length * Tangent(opposite angle).

For Length = 24 inch and opposite angle = 1 deg, then Height = 0.418922

If you have MS Excel, then you can enter a formula
Cell for Height = =Cell for Length*TAN(RADIANS(Cell for angle))

For some strange reason MS designed Excel to use RADIANS in angle functions
instead of degrees where 180 deg = PI Radians.

Dave Paine

"Joe" <[email protected]> wrote in message
news:[email protected]...
>I want to double check some wedges I've made for angled cuts. In a 24"
>long wedge, how much gain on the square corner is there for each degree of
>angle?
>
> I know the method is rattling around back there amongst the unused and as
> yet unkilled brain cells but I can't seem to come up with it.
>
> Thanks to the more mathmatically inclined.
>
> jc
>
>

Og

"Old guy"

in reply to "Joe" on 31/12/2006 10:23 PM

01/01/2007 1:03 PM

Funny story on that subject:

On of my aspiring architectural drafters was trying to make a model of his
project with a 6/12 roof slope. The roof just wouldn't work out.

After lots of investigation, I found out that he had reasoned that if a
12/12 slope gives a 45 degree angle, a 6/12 should give a 22.5 degree angle.
He just couldn't see my arguments to the contrary.

The only way I convinced him was to ask him to draw a roof slope at 24/12,
and see if it produced a 90 degree angle....

Old Guy



oe" <[email protected]> wrote in message
news:[email protected]...
>I want to double check some wedges I've made for angled cuts. In a 24"
>long wedge, how much gain on the square corner is there for each degree of
>angle?
>
> I know the method is rattling around back there amongst the unused and as
> yet unkilled brain cells but I can't seem to come up with it.
>
> Thanks to the more mathmatically inclined.
>
> jc
>
>

l

in reply to "Joe" on 31/12/2006 10:23 PM

31/12/2006 9:56 PM

In article <[email protected]>,
Joe <[email protected]> wrote:
>I want to double check some wedges I've made for angled cuts. In a 24" long
>wedge, how much gain on the square corner is there for each degree of angle?
>
>I know the method is rattling around back there amongst the unused and as
>yet unkilled brain cells but I can't seem to come up with it.
>
>Thanks to the more mathmatically inclined.
>
>jc
>
>

What do you mean by "gain" ? If you mean, for a right triangle with
angel A between the hypotenuse and a 24" side, that the "gain" is the
length of the side opposite angle X, then the change in the length of
that side is not a linear function wrt the angle X. However, it will
be equal to 24 X sine(A)

--
Often wrong, never in doubt.

Larry Wasserman - Baltimore, Maryland - [email protected]

l

in reply to "Joe" on 31/12/2006 10:23 PM

01/01/2007 12:28 AM

In article <[email protected]>,
Wilson <[email protected]> wrote:
>OOPS, maybe a little doubt would be good on this one.
>G/HYP does = SIN X, but you have two unknowns, since the base is fixed at
>24.
>G = HYP*SIN X doesn't help, since we don't know the hypotenuse.
>As Tin said, TAN X = G/24, so G = 24 TAN X.

You are right, I was mixing up my sides, the tangent is correct.


--
A man who throws dirt loses ground.

Larry Wasserman - Baltimore Maryland - [email protected]


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