RF problems in the shack part 2, success !
Well I finally solved the RF in the shack problem and it was all down to proximity.
Living in Scotland may well mean there is lots of mountains glens and lochs with plenty of fresh air and space, but not when it comes to the typical modern building plot !.
The RF problem was reduced with lower power and increasing frequency so after looking carefully at the layout of my low dipole and by low I mean a 40/80 meter dipole (half sized on 80) at only 5 metres above the ground, I decided that I had nothing to loose by trying to improve my aerial mini farm.
Now 5 metres is pretty low for a 40m dipole and almost on the ground for 80m .Of course as one support is the apex of the roof it puts the aerial only about 2 metres above the roof and all my house wiring and electrics.
So to the solution to my particular problem was to get the aerial farther away from the house and higher up...the answer to all radio amateurs dreams. The problem of course is the small garden so how do I squeeze the dipole in, well I make the horizontal dipole into an inverted V type and all at the bottom of the garden which is only 10 metres wide but 12 metres away. By increasing the height I then reduce the width of the aerial so it will fit if the feed point is high enough up, so a simple bit of maths will give you the height I need to go.
This is where the Pythagoras equation that was learnt in school and that we all said we would never use comes in to its own so here we go.
Basically it tells us that the square on the hypotenuse is equal to the sum of the squares on the other two sides so
Forgive the wee illustration above which I borrowed from www.algebra.com but as my unknown is leg B =the root of 10sqd minus 5sqd. that is 100 minus 25 = 75 the square root of which is 8.6m. So I need the feed-point at least 8.6 m high. Going up is easier than going sideways :)
So just to improve things and to keep the aerial from touching the ground or people, I have a couple of poles installed at the bottom corners of the garden at about 3 metres high so I need to get the inverted V feedpoint up another 3 metres or fold the bottom ends sideways. The increase in height at the bottom has also the effect widening the garden so opens it up the feed-point a bit.
So my RF feedback is cured and I have almost doubled the height of my dipoles so its a winning combination. As we move closer to the solar minimum getting a better aerial for the low bands is probably the way to go and by configuring the dipole as an inverted V it also acts as guy support for the top of the mast...now my only other problem is to make a rotating feedpoint as my beam is up there as well, it never ends.
Living in Scotland may well mean there is lots of mountains glens and lochs with plenty of fresh air and space, but not when it comes to the typical modern building plot !.
The RF problem was reduced with lower power and increasing frequency so after looking carefully at the layout of my low dipole and by low I mean a 40/80 meter dipole (half sized on 80) at only 5 metres above the ground, I decided that I had nothing to loose by trying to improve my aerial mini farm.
Now 5 metres is pretty low for a 40m dipole and almost on the ground for 80m .Of course as one support is the apex of the roof it puts the aerial only about 2 metres above the roof and all my house wiring and electrics.
So to the solution to my particular problem was to get the aerial farther away from the house and higher up...the answer to all radio amateurs dreams. The problem of course is the small garden so how do I squeeze the dipole in, well I make the horizontal dipole into an inverted V type and all at the bottom of the garden which is only 10 metres wide but 12 metres away. By increasing the height I then reduce the width of the aerial so it will fit if the feed point is high enough up, so a simple bit of maths will give you the height I need to go.
This is where the Pythagoras equation that was learnt in school and that we all said we would never use comes in to its own so here we go.
Basically it tells us that the square on the hypotenuse is equal to the sum of the squares on the other two sides so
 |
| half dipole length 10m (hypotenuse C) half garden width 5m (Leg A) Height I require is unknown (leg B) |
Forgive the wee illustration above which I borrowed from www.algebra.com but as my unknown is leg B =the root of 10sqd minus 5sqd. that is 100 minus 25 = 75 the square root of which is 8.6m. So I need the feed-point at least 8.6 m high. Going up is easier than going sideways :)
So just to improve things and to keep the aerial from touching the ground or people, I have a couple of poles installed at the bottom corners of the garden at about 3 metres high so I need to get the inverted V feedpoint up another 3 metres or fold the bottom ends sideways. The increase in height at the bottom has also the effect widening the garden so opens it up the feed-point a bit.
So my RF feedback is cured and I have almost doubled the height of my dipoles so its a winning combination. As we move closer to the solar minimum getting a better aerial for the low bands is probably the way to go and by configuring the dipole as an inverted V it also acts as guy support for the top of the mast...now my only other problem is to make a rotating feedpoint as my beam is up there as well, it never ends.
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