In reply to M1EYP:
If the two of you operate as a team, one could take a small tent and a small gas-powered radiant heater and the other the radio gear!
I don’t know how a SOTA reference would fit into the usual contest exchange…
73
Brian G8ADD
In reply to M1EYP:
…with the following also worked by other G SOTA ops:
LY, UA9, TF, C3, T9, UA2, UN, YL, YO.
…and don’t forget my E7… not Namibia but it counts 
73 Marc G0AZS
Indeed Roy. That is why, if I get to go out, I won’t be sitting on a frequency calling ‘CQ SOTA’. I will be hopping around answering other stations calling ‘CQ Contest’. Not very useful for chasers, but a different flavour to ‘activating’.
Brian, when in a contest, I don’t give my SOTA reference to every contest station, just make sure it get to announce it every ten minutes or so. No point giving them information they don’t want!
Marc, I hadn’t forgotten - E7 = T9
Tom M1EYP
In reply to M1EYP:
Marc, I hadn’t forgotten - E7 = T9
Ah yes sorry… the old prefix.
73 Marc G0AZS
In reply to M1EYP:
Looks like im still going to have a go and another aerial near completion built specially for the day. Sean / M0GIA
In reply to M1EYP:
Hi Tom. Interesting subject.
I do not believe that contests would be very efficient for a SOTA station on it’s own. Contest participants have not the time to listen for qrp…
But you could perhaps make an arrangement with a high-power station close to your location who can support you (outside contests), calling for new DXCC and help you with a qso, keeping the qrg clean etc. This should perhaps be done on other than SOTA frequencies.
Good luck
73
Mike
In reply to LA5SAA:
It was near completetion but i started over and now finished http://homepage.ntlworld.com/s.amesbury/M3FVB/20m%20port.htm Sean / M0GIA
Nice project Sean. Hopefully, we will both get the XYL pass-outs, and the wx next weekend. I’m hoping for some new DX DXCCs, and to get on the verge of completing Europe. I still need bits and bobs like HV and ZA.
BTW, yes there is indeed a mathematical way to calculate the length of your 45 deg radials. It is whatever the height is where they attach to the pole, multiplied by 1.414.
Cheers, Tom M1EYP
In reply to M1EYP:
Ive got the XYL’s approval and looks like Andy (M1BYH) and Greg (2E0RXX) may also join us. Its down to the WX now for me on whether its a goer.
I see its Trig your using … 1 known length and total degrees, i should have asked you first makes my way look like something early man would have done HI! Never mind i will possibly add that info to the page as an alternative to building wooden triangles. Sean / M0GIA
Possibly trig (the reciprocal of the sine or cosine of 45 degrees), but to my mind more Pythagoras (the hypontenuse of a right-angled triangle where the two shorter sides are of equal length = 1).
Tom
In reply to M1EYP:
Or a wooden triangle! Tom you know i aint got a clue on maths and the way you start spinning formulas and numbers round leaves me beyond confused. If it works dx then that will do me. Sean / M0GIA
Pythagoras:
Right-angle triangle (or set square).
Three sides, a, b and c. c is the longest side (“hypotenuse”).
Pythagoras’ Theorem is: a x a + b x b = c x c
So, to get the length of the longest side, you square the lengths of the two shorter sides, add them up, then square root the total.
So if your set square had shorter side lengths of 1m each, then the longest side would be:
1x1 + 1x1 = 2, then square root of 2 = 1.414
Hence why the radials need to be 1.414 times the height they reach off the ground, if they are to be angled at 45 degrees.
Or do you want the trig explanation…? 
Tom
In reply to M1EYP:
Or do you want the trig explanation…? Yes please Tom and i understood that then i probably wouldnt have passed my full paper had i not had the maths lessons from you. Sean
NP Sean, here we go:
You want the angle to be 45 degrees. You know the height AGL the radial will attach to (call it x). You want to know the length of the radial.
Trig needs you to name the sides of your right-angle triangle. The longest side, just like in Pythagoras above, is called the hypotenuse. The side next to the angle is the “adjacent”. The side opposite the angle is called the “opposite” (funnily enough).
In this case, it doesn’t matter which way round they are, because the angle is 45 degrees and therefore the opposite and the adjacent are the same length as each other.
However, let’s consider the 45 degree angle to be between the radial and the pole. So the pole is the adjacent. The radial is the hypotenuse. We know the height of the adjacent - x, the height AGL where you connect the radial.
The 3 trig functions are:
sine = opposite divided by hypotenuse
cosine = adjacent divided by hypotenuse
tangent = opposite divided by adjacent
We have the hypotenuse (the radial) and the adjacent (the height AGL) involved in our question, so we select the cosine ratio to use.
We want to know the hypotenuse (length of radial).
So we rearrange the function to be:
hypotenuse = adjacent divided by cosine
so we get:
radial = height we connect to divided by cos45 (=0.707)
dividing by 0.707 is the same as multiplying by 1.414
This is all a lot easier than it looks here - if I write it down for you as formula triangles, you’ll find it easy. Remind me.
By the way, why not register a new account on here, then you can be M0GIA.
Tom
In reply to M1EYP:
Ok to stop any confusion im M0GIA and straight back to the math i thought i had seen the figure 1.414 then .707. Thats RMS as i was taught at college but never knew it had anything to do with triangles but this is now going beyond the subject of this thread.
When i see you next i might just pick your brains on something not related to radio. Sean with a new account!
Nonsense Sean, it’s all totally on-topic. 1.414 is the square root of two. 0.707 is one divided by the square root of two (the reciprocal), also equal to half of the square root of two. It only applies to isosceles right-angle triangles, ie triangles with two 45 degree angles and a 90 degree angle. But in all such triangles, the lenth of the hypotenuse is root 2 (1.414) times longer than one of the other sides.
Tom
In reply to M1EYP:
Your really going there now and what i want to know is how its connected to EMF focus and Orgone if it truley exists and some experimenting i was working on a few years back. F sharp was noted to be connected but i thought this was a low frequency then i might have been missled? Any way the vitamin tablets did become too much to stomach and the aroma from them flooded the house at which point this experimenting stopped. Sean
Now it’s you that’s lost me Sean!
Tom
Collating the contributions above, and adding a few new ones from the CQWW this weekend, the collective list of DXCCs worked by SOTA activators in G is:
4X, 5B, 9A, C3, CT, CT3, DL, EA, EA8, EI, ER, ES, F, G, GD, GI, GM, GW, GU, HA, HB, I, K, LA, LX, LY, LZ, OE, OH, OK, OM, ON, PA, RA, S5, SM, SP, T9, UA2, UA9, UN, UT, VE, YL, YO, YU, ZB
Total = 47. Any more for any more?
Tom M1EYP
I know that TF has been worked from 2 different G summits during activations Tom. Can supply the precise info if you need it.
73, Mike G4BLH