January 6, 2002 01:51
OK, here it is. The GPS method looks to be sufficiently accurate
to meet Tom's needs. This is pretty lengthy as I want to show how
I got to the solution so that if I have made a mistake others will
be able to point it out.
To use the method, you need a reasonable quality consumer grade
handheld GPS unit and the ability to do very simple trig
A description of the problem is followed by a discussion on GPS
accuracy, the required steps to lay out the rectangle, the
equations required for the calculations and a worked example. The
calculations could all be done very easily with a spreadsheet. If
enough people ask me to create a spreadsheet to do this, I will.
How many is enough? How about 20? Suggestions as to a site from
which people could download the spreadsheet would also be useful.
If 50 people want it, I'll create a little Visual Basic program
you can just plug the numbers into and it will give you the
answers as well as draw a little diagram so that you can see if
the numbers make sense. No need to mess with spreadsheets and no
need for Excel. Not 'til after contest season though.
Tom needs to be able to locate on the ground the corners of an 800
ft x 400 ft rectangle to an accuracy of a few dozen feet. The
rectangle long sides will point in a specified direction. The
accuracy of the bearing is less significant than the accuracy of
locating the corners. Normal sighting and tape measure methods
(at the non-surveyor level) are difficult and time consuming due
to hilly ground and obstructions.
He doesn't need to locate the corners of the rectangle at precise
coordinates, he just needs to lay out a rectangle to meet the
above specs. Consequently, absolute accuracy of the GPS unit is
irrelevant and relative accuracy from one location to another
needs only to be within a few dozen feet.
GPS ACCURACY. IS IT GOOD ENOUGH?
I have an old Garmin GPS II+ handheld unit. It will receive 12
satellites and will perform averaging to make position
measurements more accurate. I conducted a test with this unit
over a period of about a day to see how much the reported
coordinates of a fixed position varied. In order to maximize the
number of satellites picked up and still stay dry here on the wet
coast I put the GPS antenna on the roof and dropped the coax
through a conduit into a room below. This is a proper GPS remote
antenna with preamp. I recorded the position of the antenna as
reported by the GPS at intervals. The idea being that if the GPS
always reported the same coordinates within a few dozen feet then
it would meet Tom's needs. Measurement protocol and results
You might want to try the same thing with your GPS before using it
to lay out an array.
The GPS II+ shows something called EPE on the satellite display
page. This is Estimated Position Error in whatever distance units
you have selected. I used feet. What does it mean? Who knows.
It's listed twice in the manual index but nothing about it on
When averaging while marking a waypoint the unit shows a number in
distance units called FOM which stands for Figure Of Merit. This
is a value which Garmin says "reflects the estimated accuracy of
the averaged position". No idea how that relates to accuracy
1. Turn on the GPS and watch the satellite display.
2. When all visible satellites have been acquired write down the
time, number of acquired satellites and EPE.
3. Press the MARK button to capture the position.
4. Select the AVERAGE option.
5. When the FOM stabilizes write down the coordinates and FOM
6. Check that the number of acquired satellites hasn't changed.
(Didn't check that they were the same satellites, just the same
number of them.)
7. If the number of satellites changed during the measurement
process repeat steps 2 to 5. Otherwise goto next step.
8. Turn off the GPS. (No need to actually save the waypoint.)
If you want to try this for yourself, don't worry if your unit
doesn't give you EPE or FOM. I only included them here in case
they mean something to someone.
What I didn't do.
I didn't record short term changes in position. e.g. write down
the Lat and Long every 10 seconds for 2 minutes. There are short
term changes but I didn't see any that would invalidate the
Degrees aren't shown because they never changed.
Integer portion of the minute values aren't shown because they
Decimal fraction portion of minute values are shown.
Date Time # of EPE LAT LONG FOM
PST Sats ft minutes minutes ft
Dec 29 1926 10 11 0.187 0.863 10.1
2055 9 13 0.185 0.862 12.6
2150 10 13 0.184 0.864 12.3
Dec 30 1003 8 15 0.185 0.863 13.8
1213 10 13 0.185 0.864 12.4
1420 9 11 0.184 0.857 10.6
1604 7 18 0.185 0.858 16.2
1916 10 11 0.186 0.861 10.2
2115 10 12 0.184 0.861 10.9
The difference between the lowest and highest LAT numbers is
0.003'. This translates into 18 ft.
The difference between the lowest and highest LONG numbers is
0.007'. At my Latitude one minute of longitude is 3989 ft so
0.007' corresponds to 28 ft.
Both of these numbers are well within Tom's accuracy
requirements. Therefore, using a GPS II+ or equivalent to
establish the positions of the rectangle corners is a viable
PROCEDURE TO LAY OUT RECTANGLE CORNERS
1. Decide on the lengths of the rectangle sides.
2. Label the corners of the rectangle A, B, C and D in a
Lines AB and CD are the long sides.
Lines BC and DA are the short sides.
3. Determine the desired bearing of the long sides.
i.e. when you stand at A and look at B what is the angle
between line AB and True North measured clockwise from North. (If
B is due west of A the angle is 270 degrees, not 90)
4. Decide where on the ground you want to locate point A.
Make sure there are no overhead obstructions like trees which
will limit the number of satellites you can acquire. The more
satellites, the greater the accuracy.
5. Make a rough estimate of the locations of Points B, C and D
and check for overhead obstructions at those points.
If there are any, relocate A to suit.
6. Beg, borrow, rent or steal a hand held GPS unit which will do
averaging of position readings. My old GPS II+ from Garmin will
do this. The more channels it will receive the better this
averaging will be (well, anyway, it oughta be. I don't actually
know.). My unit will receive 12 channels. Some other models of
that vintage would receive only half that number. The actual
number of useable satellites at any time will generally be less
Anyway, when marking a position using averaging my unit claims
to be, typically, within 12 ft or so of being right when there are
lots of satellites. The actual claimed error depends on the
positions of the satellites and how many are in view. What
follows is based on my unit as I have no experience with any
7. Take the GPS unit to Point A and record the position on the
GPS using averaging, i.e. create a waypoint. While you're there,
drive a stick into the ground.
8. Calculate the lat and long of the other 3 corners (B, C and D)
of your rectangle from a knowledge of which way you want the array
to point and how far apart one corner is supposed to be from the
next one. How do you do this? See the formulas and worked
9. Get out the GPS again and enter 3 new waypoints using the lats
and longs of the 3 corners you just calculated. Label these B, C
and D (clockwise from A). In the example below sides AB and CD
are the long sides and BC and DA are the short sides.
My unit will allow you to enter Lat and Long to the nearest 0.001
minute which translates to 6 ft. Less than 6 ft for Longitude.
10. Set the GPS to go to one of the waypoints. Go to the
and drive in a stick.
11. Repeat for the other 2 corners.
12. You're done.
The example will use a rectangle of 800 ft x 400 ft. The corners
are labelled A, B, C and D in a clockwise direction. Sides AB and
CD are 800 ft and sides BC and DA are 400 ft. The long dimension
points ESE (112.5 deg clockwise from True North). i.e. if you
stand at A and look at B you are looking in the ESE direction. In
what follows this bearing or angle is called NAB. (Difference
between N and line AB) Corner A of the rectangle is at 33 deg
4.450'N and 84 deg 3.533'W.
TO FIND POINT B
1. Calculate the position of B relative to A in feet.
This is done in 2 pieces, how many feet E or W of A (call this
distance Bx) and how many feet N or S of A (call this distance
By = AB x cos(NAB) (Northern hemisphere)
Bx = -AB x sin(NAB) (Western hemisphere)
signs are reversed in the other hemispheres. (I think. Haven't
By = 800 x cos(112.5) = 800 x -0.383 = -306 ft
Bx = -800 x sin(112.5) = -800 x 0.924 = -739 ft
The minus signs mean B is south of A (latitude is less) and east
of A (longitude is less).
2. Convert By and Bx values from feet to degrees lat and long.
1 minute of latitude = 6,080 ft (1 nautical mile) no matter where
1 minute of longitude = 6,080 ft (1 nautical mile) only on the
equator. Everywhere else it is less but you can easily calculate
it for your latitude as follows.
feet/minute of longitude at latitude X = 6080 x cos(X)
So, 33 deg 4.450' lat = 33 + 4.450/60 = 33 + 0.0742 = 33.0742 deg
ft/min of long at lat 33.0472 deg = 6080 x cos(33.0742)
= 6080 x 0.838
= 5095 ft
i.e. 1 minute of longitude at 33.0472 deg lat is 5095 ft
Now calculate the difference in lat and long between A and B.
Call these numbers RelLatB and RelLongB. (Relative lat and long
between B and A)
RelLatB = By/6080 = -306/6080 = -0.050 min Lat from A
RelLongB = Bx/5095 = -739/5095 = -0.145 min Long from A
i.e. B is 0.050 min south of A and 0.145 min east of A.
3. Calculate the actual coordinates of B
Lat B = Lat A + RelLatB = 33 deg 4.450'N - 0.050'
= 33 deg 4.400'N
Long B = Long A + RelLongB = 84 deg 3.533'W - 0.145'
= 84 deg 3.388'W
TO FIND POINT D
1. Calculate the position of D relative to A in feet.
Because side AD is at right angles to side AB, the formulas are
Using similar notation as before
Dy = -AD x sin(NAB) (Northern hemisphere)
Dx = -AD x cos(NAB) (Western hemisphere)
signs are reversed in the other hemispheres. (I think. Haven't
Using the same example as before
Dy = -400 x sin(112.5) = -400 x 0.924 = -370 ft
Dx = -400 x cos(112.5) = -400 x -0.383 = 153 ft
So D is 370 ft south of A and 153 ft west of A
2. Convert Dy and Dx values from feet to degrees lat and long.
RelLatD = Dy/6080 = -370/6080 = -0.061 minutes
RelLongD = Dx/5095 = 153/5095 = 0.030 minutes
i.e. D is 0.061 min south of A and 0.030 min west of A.
3. Calculate the actual coordinates of D
Lat D = Lat A + RelLatD = 33 deg 4.450'N - 0.061'
= 33 deg 4.389'N
Long D = Long A + RelLongD = 84 deg 3.533'W + 0.030'
= 84 deg 3.563'W
TO FIND POINT C
>From the geometry
Cy = By + Dy
Cx = Bx + Dx
Cy = -306 + (-370) = -676 ft
Cx = -739 + 153 = -586 ft
Here's an opportunity to check your work so far. If the numbers
are right then they represent the corners of a 400 ft x 800 ft
rectangle. If it is a rectangle then the lengths of the diagonals
are equal so length AC must = length BD.
AC = sq root(Cx*2 + Cy*2)
BD = sq root((Bx-Dx)*2 + (By-Dy)*2) (The *2 notation means
So, lets try it
AC = sq root((-586)*2 + (-676)*2) = sq root(343,396 + 456,976) =
= 894.635 ft
BD = sq root((-739-153)*2 + (-306-(-370))*2) = sq root((-892)*2 +
= sq root(795,664 + 4,096) = sq root(799,760)
= 894.293 ft
I guess an error of 4 inches is acceptable
The length of the diagonal of an 800 ft by 400 ft rectangle = sq
= sq root(640,000 + 160,000) = sq root(800,000)
= 894.427 ft
This result is very comforting
RelLatC = Cy/6080 = -676/6080 = -0.111 min Lat from A
RelLongC = Cx/5095 = -586/5095 = -0.115 min Long from A
LatC = LatA + RelLatC = 33 deg 4.450'N + (-0.111')
= 33 deg 4.339'N
LongC = LongA + RelLongC = 84 deg 3.533'W + (-0.115')
= 84 deg 3.418'W
You don't actually have to go through all the Cx, Cy, etc.
calculations because, from the geometry
Lat C = Lat B + RelLatD
= 33 deg 4.400'N + (-0.061') = 33 deg 4.339'N
Long C = Long D + RelLongB
= 84 deg 3.563'W + (-0.145') = 84 deg 3.418'W
So, the coordinates of the 4 points are:
A 33 deg 4.450'N 84 deg 3.533'W
B 33 deg 4.400'N 84 deg 3.388'W
C 33 deg 4.339'N 84 deg 3.418'W
D 33 deg 4.389'N 84 deg 3.563'W
Whew! This was way harder than I expected it to be. A surveyor
could have laid out ten of these rectangles in the time it took me
to derive all this, hills and obstructions notwithstanding.
The reason for the difficulty is that in convential Cartesian
coordinates numbers on the x axis get more positive as you move
further to the right. In the Western hemisphere, longitude values
get more positive as you move to the left (west).
Also, in conventional polar coordinates angles are measured from
the x axis and increase as you go counterclockwise. When pointing
our beams we measure angles from the y axis (north) and the
numbers increase as we go clockwise.
These formulas should work in all four quadrants, i.e. for any
value of the angle NAB.
Please note that the error checking formulas shown won't work once
you have converted from feet to lat and long. They will work if
you correct for the fact that 1 min of long is a different number
of feet, depending on your lat.
If anyone finds any errors in this, I'm not sure I want to know.
However, I guess you better tell me.
Let me know if you want to put this up on a web site. I'll
rewrite the preamble a little so it doesn't rely on the thread for
Tom - I hope I didn't put Point A in your front room.
N4GI for casting doubt on GPS accuracy, thus causing me to test
WA2MOE for pointing out that the distance of 1 minute of
longitude varies with latitude.
K5RA for giving me the formula for calculating the distance of
1 minute of longitude at any latitude. I checked this
independently and agree with it.
KK6T for giving me a link to interesting GPS info
73 and sure hope this is useful to someone.
Jim Smith VE7FO
Tom Rauch wrote:
> Thanks for all the good suggestions, I'm still digesting all the
> information. It looks like some sort of really good sighting compass
> would work if I stay away from the barbed wire fences.
> What I need to do is first lay out long wide-spaced antennas.
> Typical antennas are 800 feet long in a straight line, and 400 feet
> apart broadside. I have to get the endfire stagger and length within
> a few dozen feet to prevent pattern degradation, and that is tougher
> than it seems when the ground is a little hilly and the view is
> obstructed. The exact direction is less important than making an
> accurate rectangle.
> Using a regular cheap small sighting-type compass (from K-Mart
> sports) to make a rectangle only gets me within maybe 70 feet and
> it is very difficult to do that well. Right now it takes me hours to get
> the layout close, mostly because I have obstructions from trees
> that block the view so I have to rely on the compass. It takes
> longer to lay an array out than to install and connect it!
> As a secondary use, I thought I could define my property lines
> between survey posts that have very large separations. That way I
> can be sure my antennas don't run into hostile territory, since the
> old wire fences are off by hundreds of feet!
> 73, Tom W8JI
This list is sponsored by the new eHam.net Store. When you buy products
like ICE filters, Array Solutions StackMatch, or M-Squared Antennas from
the eHam Store for the same price you pay direct, a portion of the sale
price goes to support this list. Check it out at http://www.eham.net.
FAQ on WWW: http://www.contesting.com/FAQ/towertalk
Administrative requests: towertalk-REQUEST@contesting.com