Well, is this the first casualty of the move from one web host to another. The files are in place somewhere but the last post in this Blog is dated December 2002 ... keep watching! DW

I'm back ... did you miss me? I've been off line for much of this week. Changed Web Host and took a trip to a seminar in London and then decided not to upload the site again until I had broadband installed at home ... broadband arrived today and here we are again! Welcome back! DW

Can you imagine a 6 foot 2 and a half inch 93 kilogramme frame slithering across the dance floor in a meaningful way to the Salsa rhythm? Neither could the young lady who tried to learn the basics of Salsa from me following on from my own one hour lesson of earlier in the evening! Went to a very entertaining Salsa evening at Oxford Town Hall last night. It started with a lesson for the uninitiated like me. Of course, I thought I did well and was ready to do a John Tavolta all over the knot end. My prospective partner had always been prepared (warned seems the operative word with hindsight) to learn from me but when we hit the floor it took her the length of time it takes light to travel one metre to decide that my Salsa skills were already worse than hers and she knew nothing. We then stood and admired the rest of the people on the dance floor for a while then made our excuses and left! Dima is a natural dancer and with his partner Joanne he did a lot better than me: at least Jo was prepared to stay on the dance floor with him for the entire evening! Mrs W can't dance at the moment because of a sharp pain in her hips. Dima is ballroom trained and was inspired by the end of the evening to ask Jo is she wanted to take Salsa and other dancing lessons with him and she gave him a tentative yes: hope it works out for them. Isn't it sickening to see people with such rhythm and talent only for yourself to realise that God didn't bless everyone with the same genetic complement?? It was hot and sticky there too so when we got home, at just after 12:30 am Mrs W ordered : showers for EVERYONE! Sir, yes sir!! Son Andrew answered the call and provided the solution to a sticky maths problem for us: all to do with circles, tangents and isosceles triangles. Here's the solution ... you can imagine the question! Myanswer is... (2y - x - 90)/2 Triangle AOC is a isocoles triangle, as line OA and OB are equal to the radius, therefore angle OBA must be (180 - x)/2. Angle CBO must then be 180 - ((180-x)/2 + y ), giving (180+x-2y)/2. Triangle OCB is also isocoles as line OC is equal to the radius, so angle OCB also equals (180+x-2y)/2. Angle OCT is also a right angle as line CT is tangential to the circumference. BCT must then equal 90-OCB, which is 90 - ((180+x-2y)/2), which equates to (2y-x-90)/2 DW