Re: Formula for figuring the MOS of a generator.

I not only talk to myself and answer myself i sometimes come up with an objective
third opinion. Talk about math. Well he it goes Maybe Joe should include this
under MOS.

This is a formula for finding the MOS of a given generator. It work even if you
don't understand it. You need a scientific calulator though that can do 1/x or X-1
(the negative one is an exponent). These two are the same thing and will give you
the antilog of the log of the interval generator we will be using.

Lets take for example the the 3/2 and see what the MOS are. first we find the log.
.5849625007
now we are going to find what is referred to as the the Zig-Zag pattern which we can
use on the scale tree but is't nessicery.. One more thing before we start. that is
the limitations of calulators. We will be subtracting numbers from the left which
means after 9 steps our orgininal data of 10 numbers (.5849625007) will be gone. the
last number7 is questioable as being appox. so we only go 9 places. You can go
further but you are calulating calulator artifacts not the real thing. To go further
you need to get a bigger machine.
now we find the antilog (1/x or the X-1) this is 1.709511291 we subtract the
number to the left of the decimal point, place it in a column and repeat this
process 8 more times.

example from beginning
.5849625007 1/x=
1.709511291 -1= 1 l
.7095112914 1/x=
1.40942084 -1= 1 r
.40942084 1/x=
2.442474596 -2 = 2 l
.442474596 1/x=
2.260016753 -2= 2 r
.260016753 1/x=
3.845906042 -3= 3 l
.845906042 1/x=
1.18216439 -1= 1 r
.18216439 1/x=
5.489547094 -5= 5 l
.489547094 1/x=
2.042704393 -2= 2 r
.042704393 1/x=
23.41679441 -23 23 l

that's are far as we can go! Now we either go to the scale tree or construct one. We
put 0/1 and 1/1 (or find this on the scale tree) at the top of a page widely spaced
to insert data in between. using what Wilson calls "Freshman Sums" (adding the
numerators, then the numerators) we move toward the left of 1/1 (always the starting
place). This means we add the number by freshman sums that to the left. Its always
the diagonal directly above. we move 1 to the left being our first number 0/1+1/1 =
1/2. we place this somewhat below with lines diagonally down from the 0/1 and 1/1
(simile throughout)
then 1 to the right (adding the 1/1+2/1 =2/3
then 2 to the left (add 2/3+1/2= 3/5 then +1/2 again=4/7
then 2 to the right (4/7+3/5=7/12, 7/12+3/5=10/17
then 3 to the left (10/17+7/12=17/29, +7/12=24/41, +7/12=31/53
then 1 to the right (31/53+24/41=55/94
then 5 to the left (55/94+31/53=86/147,+31/53=117/200,+31/53=148/253,
+31/53=179/306, +31/53=210/359)
then 2 to the right (210/359+179/306=389/665, +179/306=568/971
then 23 to the left you can do this one. see http://www.anaphoria.com/ST10.html as
a main section of this diagonal down.
The denominator is the MOS the numerator is number of steps of the generator.

Feedback sought!

Kraig Grady wrote:

> I realize that Erv fomula for finding the MOS of a given
> Generator is not up so I am going to put it up in a seperate post so it does'nt
> get lost in the middle of this!

-- Kraig Grady
North American Embassy of Anaphoria Island
http://www.anaphoria.com

In-Reply-To: <934022737.27733@onelist.com>
Back in digest 273, message 15 Kraig Grady gave a formula, and asked
for feedback on it. Well, sorry this is a bit late coming, but yes
it looks simple and effective. I've managed to tell my computer how
to go left and right along the tree, and so it can now reproduce the
zig-zag. I'll try and get the code into an applet for my website
sometime.