"...daves report roughly midway down the page
<
http://tinyurl.com/6zgwc> "
Ah. A key message I apparently overlooked while out of town.
That answers some questions I asked in a later message.
"...limit our overall degree of certainty
to 2 decimal places..."
Yep, at best 2 places.
My programs use about 12 digits total
for deg min sec and other calculations.
BUT, it can't make the output better
than the output.
"..the submillimetric exactitude of your solution
here below seems to indicate that you dont agree..."
I believe we are on the same page and same song.
One way to try to preserve the 2 places input
is to do the arithmetic to at least one more
and then round to 2 at the end.
The program just shows all the digits.
> But, how to find the north-south distance from
> a 3rd point to an intermediate point on the line
> isn't obvious.
"ok but i still think it is obvious
at least within the above stated parameters of exactitude
it seems entirely reducible to a simple ratio computation"
True for the short distances in this problem,
but for the general case where the points
are a lot further apart plane trig stops working
and spherical trig has to be used, as in the formula
(which works for short and long if one is careful
to account for the special cases).
> ==========================
>
> RCMcC
> Angles and Distance greatly distorted.
"what do you mean specifically by the above comment?"
Just that it's hard to draw a triangle
that's 3212 meters long and only 0.08 meters tall.
>
> [194] -- 3212 m -- I -- [195]
> \ | /
> 2946 m 267.048 m
> \ /
> [LSAW 5]
>
> ==========================================
"> Given (all NAD-27 converted to NAD-83 with CORPSCON )
>
> Monument #194 = Point 1 = Lat1/Long1 = N 48.999056º W
117.072683º
> Monument #195 = Point 2 = Lat2/Long2 = N 48.999215º W
117.028771º
> LSAW#5 plaque = Point 3 = Lat3/Long3 = N 48.999197º W
117.032421º
"but i think your 6th digit may be questionable
for points 1 & 2"
Maybe so. I used all the NAD-27 digits
to feed into the standard CORPSCON translator to NAD-83
and then rounded that output off to 3 digits
to try to keep as much as possible until after
all the number crunching.
The final result after all calculations for Point I
on the boundary is only good to 2 digits - at most.
Then one should also round points 1 and 2.
"...crediting such exactitude as you are delivering..."
The math is just manipulating a model of physical reality.
When physical reality is different from all the
digits put out by the computer,
physical reality rules
and you try to correct the model.
"...someone who understands all this ..."
Gotcha fooled. :)
I'm definitely not a pro at this stuff,
just someone who has fiddled with it in spare
time for a long time trying to keep the models
(borrowed from the _real_ pros) as accurate as possible
when crunching the numbers through a computer.
I've studied a lot of sloppy programs
where someone encoded a formula from a textbook
into a computer program and had no idea
what its limitations were and how the limitations
of the computer arithmetic affected it.
The GUIs looked great, though.
A few years ago I was reading a program
where I found that the programmer had
twice typed in the value of pi as
3.141592818
where the three digits "818" after the "2" are wrong.
Fortunately, in his particular case it made
no real difference, but if he were trying
to calculate distances down to inches,
as we are attempting here - no good.
I guessed that he may have obtained
the wrong value of pi from the MS Windows calculator.
We are going out of town tomorrow and it may be
Monday before I get to check in again if I miss
in the morning.
Have fun.
Cheers, 73,
Ron McC.
w2iol@...
Ronald C. McConnell, PhD
WGS-84: N 40º 46' 57.6" +/-0.1"
W 74º 41' 22.1" +/-0.1"
FN20ps.77GU31 +/-
V +5058.3438 H +1504.2531
http://home.earthlink.net/~rcmcc
There are 10 kinds of people.
Those who understand binary arithmetic
and those who do not.