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11-10-2012, 09:15 AM
Guys, I am installing a ground mounted vertical antenna for my Ham Radio setup. I would like to put 60 wire radials around it in the ground about 1/2" . If I draw a circle around the antenna base 33 feet in radius, how far apart will each wire be from the other at the 33' point? I'm thinking the trig is WAY (yes I'm shouting) over my head and maybe someone can do an example in cad and lift a dimension for me. I'm guessing it's going to be about 5" from each other at the outer edge of the circle.

It is easier to lay the radials starting at the outer edge and working towards the center of the circle where the antenna mount is as that many radials start to step on one another!

Ray......

PS: Putting up a KLM 40 meter 2 element linear loaded beam and a KLM KT36XA 6 element tri-bander with 10 db of gain this Monday. With a KW amplifier I should be heard on the ham bands! Should net me a "big-a-deal" signals as the Spaniards are fond of saying! :D The vertical is being installed at the cottage as it is less disruption to the neighbors.

Haven't been around much, nice to see everyone getting along well :D

rythmnbls
11-10-2012, 09:28 AM
(66 x PI)/ 60 would get you pretty close which works out at about 3 feet 5.4" apart at the circumference.

Regards.

Steve.

DFMiller
11-10-2012, 09:32 AM
A simple approach might be divide the circumference by the number of radials..
So that would be 66 times 3.141519
Or abou 207 feet
Divide by 60

I would read this website before I put all the wire up.
http://www.antennasbyn6lf.com/
You might go with elevated radial and save some wire.
Good DX
Dave
VE7PKE

John Stevenson
11-10-2012, 09:35 AM
Simple.
Formulae is Pi*D so 3.1416 x 66 = 207.3456 feet

207.3456 feet x 12 = 2488.1472"

2488.1472" / 60 = 41.469"

However this is measured on the circumference and the chordal length from point to point is less at 41.45"

Lew Hartswick
11-10-2012, 10:21 AM
33 ft x Tan of 6 deg is easier for me. :-)
...lew...

Barrington
11-10-2012, 01:30 PM
To avoid accumulating errors, other numbers that may be useful for laying out the radials might be the distances spanning multiple endpoints.

The straight-line distance between non-adjacent endpoints is:

66' * sin (3 * N)

- where N is the number of endpoints spanned.

e.g. the distance from endpoint #1 to endpoint #6 will be: 66' * sin (3 * 5) = 17' 1"

So:-

1 -> 6 : 17' 1"
1 -> 11 : 33' 0"
1 -> 16 : 46' 8"
1 -> 21 : 57' 1 7/8"

etc.

Cheers

.

11-10-2012, 01:40 PM
Surely if you have CAD such as AutoCad, it will do it all for you?
Max.

JRouche
11-10-2012, 02:14 PM
Hey YOD.. Solly,I cant help cause math is my weak subject. But all the smart folks gotcha going. I just wanted to say HEY! Haven't seen yer Sig in awhile. JR

Paul Alciatore
11-10-2012, 02:39 PM
Like the others said, CAD is not needed. Several have suggested dividing the circumference of the 66 foot diameter circle by 60. This is not an exact answer, but is probably close enough for your purposes. You may have to do a bit of adjusting when it is done. Better yet, you could get four starting points by laying out a diameter in a straight line across the center which will give you two starts. Then lay out another diameter at right angles to that one by using equal distances from the two end points on the first diameter. That will give you four starting points at 90 degrees apart. Now, use the 41.5 inch number from several of the above posts to layout eight points in each direction from each of these four starting points. The eighth points from adjacent pair of starting points should coincide and this will give you an idea of the amount of error you have.

If you want a mathematical exact calculation, you need to find the chord between two adjacent end points instead of the arc length. To do this you need to use a half angle calculation and then double the result. Since you have 60 radials they are 6 degrees apart. The half angle is then 3 degrees. You use the sine of 3 degrees as follows:

Half chord (in inches) = 33 x 12 x sin(3)

Half chord (in inches) = 20.7250"

Chord = 2 x Half chord = 2 x 20.7250"

Chord = 40.4500"

As you can see, this only differs from John Stevenson's 41.469" by 0.019". His figure is probably close enough.

John Stevenson
11-10-2012, 03:21 PM
Paul, Carry on reading, I did continue and state the chordal length at 41.45.

Rustybolt
11-10-2012, 04:54 PM
If you have a cad program just do a radial array giving the radius and the number of elements. You can then access the chordal distance by dimentioning the distance between two adjacent elements

oldtiffie
11-10-2012, 05:47 PM
Borrow a theodolite, set it over the centre point, sight the first point, zero the transit, get a friend to stretch the tape (no "snags"), mark the 30 foot mark on the tape (tension the tape correctly) with a "peg/marker".

Set the transit to 6 degrees and repeat.

Set at each 6 degree increment until you reach 360 degrees (zero) and the transit should be sighting your original mark/er (at zero degrees) as a check.

You can "tilt" the theodolite to mark out say the 10 or 15 foot points along each radial as you go as well (as a check to see the radials are set out straight).

No CAD, no maths, no trig, almost no time at all and the minimum of error and effort.

Paul Alciatore
11-10-2012, 06:07 PM
Sorry John.

Paul, Carry on reading, I did continue and state the chordal length at 41.45.

11-10-2012, 06:39 PM
Simple.
Formulae is Pi*D so 3.1416 x 66 = 207.3456 feet

207.3456 feet x 12 = 2488.1472"

2488.1472" / 60 = 41.469"

However this is measured on the circumference and the chordal length from point to point is less at 41.45"

Thanks to Sir John, Earl of sudspumpwater and mentor to the mathematicly challenged worldwide for giving me an answer I can completely understand. Had I thought about the problem for a while longer I might have come up with that answer on my own as it makes complete sense to me.

AND, a very special thanks to all of you who continued answering this question even though it had been answered. Reading your words made me feel really smart save one mention. If I had a chordal, would I set it up on my lathe, mill or shapper? It's likely I have a few chordals laying around I'm just not sure what they are called on this side of the pond?

I will cut me a stick 41 1/2 inches and run strings from each end out 33 feet to my antenna mount and mark it off. From an RF point of view it isn't that critical they be spaced so evenly but at the price of copper wire these days it will keep me from getting more radials on one side then on the other which would effect the antenna pattern to "some" effect. It's just as easy to get it right as not.........well, as long as I can sign into the forum it is :D

OldTiff, thanks for the suggestion but my wife took the theodolite we got as a wedding present to the pawn shop and hocked it for \$50.00 to buy some new shoes. You can imagine my dismay when I found out she threw out the old pair that we had been sharing all these years.

JRoach, good to see you as well. I do a lot of lurking these days. Recently used my shop to rebuild a 40 meter 2 element linear loaded beam that was given to me damaged. Also made parts for a 6 element tri-bander on 32 foot boom. It all gets in the wind this Monday with help from my radio club guys.

73's (as they say in ham speak)
Ray.........( NV2A )

11-10-2012, 07:54 PM
Like the others said, CAD is not needed.

.

??Cad would give everything in just a few seconds?, No math needed.
Max.

h12721
11-10-2012, 08:27 PM
By the time You did asked the question and typed it in her , waited for the answers and read them all I would have the job don . On 40 meters it would not mean a hood if it would be 30, 40 or 60 or what ever. Also it could care less if there are 33 or 133 long But in any case lay on radial out lay the next one 180° to it keep on dividing in half till you get to it. Job don. You on the Air with so so results
sb 220 forsale
Hilmar, WB2nec

11-18-2012, 07:38 AM
By the time You did asked the question and typed it in her , waited for the answers and read them all I would have the job don . On 40 meters it would not mean a hood if it would be 30, 40 or 60 or what ever. Also it could care less if there are 33 or 133 long But in any case lay on radial out lay the next one 180° to it keep on dividing in half till you get to it. Job don. You on the Air with so so results
sb 220 forsale
Hilmar, WB2nec

Actually Hilmar, the requirements for proper radial field under a ground mounted verticle are much different then for a roof mounted installation. For a roof mounted installation you would be more correct but for a ground mounted system more is better and good spacing helps keep the antennas pattern equal in all directions. Is it as critical that 1" would make a difference, heck no but it's just as easy to get it right as to just drop a bunch of wires on the ground. My guess is if someone tried to lay 60 radials in a nice array without measuring they would end up with a mess. 73's Hilmar

11-18-2012, 09:28 AM
Guys, I am installing a ground mounted vertical antenna for my Ham Radio setup. I would like to put 60 wire radials around it in the ground about 1/2" . If I draw a circle around the antenna base 33 feet in radius, how far apart will each wire be from the other at the 33' point? I'm thinking the trig is WAY (yes I'm shouting) over my head and maybe someone can do an example in cad and lift a dimension for me. I'm guessing it's going to be about 5" from each other at the outer edge of the circle.

It is easier to lay the radials starting at the outer edge and working towards the center of the circle where the antenna mount is as that many radials start to step on one another!

Ray......

PS: Putting up a KLM 40 meter 2 element linear loaded beam and a KLM KT36XA 6 element tri-bander with 10 db of gain this Monday. With a KW amplifier I should be heard on the ham bands! Should net me a "big-a-deal" signals as the Spaniards are fond of saying! :D The vertical is being installed at the cottage as it is less disruption to the neighbors.

Haven't been around much, nice to see everyone getting along well :D

when I was active g4ijx we used to make the radials different lengths if that vert is for more than one band make them 1/4 wave for each band dispersed evenly round the base just my thought

I uses to use a 3 kw amp ( illegal in the UK we were limited to 400 pep ) on top band through to 10 meters , with a about 1kw on 432 megs

73's
Stuart station now dismantled