Everything posted by MikeGroves
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Service update...........
yuh and torque up everything to the lb as in the manual ;) very very very important, you don't want anything coming undone or breaking under to much pressure.
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Service update...........
Jeez looking good, I did this in the summer and that was bad enough, god knows what you are going through in this tempreture :D i've put off my cambelt change until the spring at least,
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Any recommended Vodafone mobile companies?
three are rubbish for support (all indians) and coverage is sketchy at best of times, my dad has cancelled his contract and my girlfreind is in the process of closing hers
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Any recommended Vodafone mobile companies?
orange will give you cross network minutes now, vodaphone custoemr service is rubbish, orange is ace :)
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Nearly
??
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Essex Curry Pics
LMAO :D
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Essex Curry Pics
From Left to right JD's-Z, Russ, my mate duncan, MikeGroves, russ's mate (matt??), MrP Me, Deve8uk MrP, and Takemetothepub (constipated it appears)
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what are my options
they told me they did national when i spoke to them, phone them, you can always park up on my drive whilst they fit it as well mate if they won't come out, its not a problem, or maybe get them to ship the glass to you, the glas was about 186quid i believe
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what are my options
http://www.windscreensdirect.co.uk/ £200 fitted for my car, it was uk but they could get a brown tint they told me, that included the clips to fit and the bloke to drive to me, he actually thought 200 was expensive so don't act surprised when he offers you that prices ;) hope that helps, it was the cheapest i could find new glass for
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**URGENT** Help with removing the Crank Pulley Bolt....Please
lmao there is a pattern emerging with you in this thread :D:D
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Grrr scally w4nkers
thats shitty :( doubt they got in, they never close the door ****ers how much are you paying for the lockup?
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just had my car cleaned
they appear to have taken all the white off mate :D looks good, any plans for this one?
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Essex Curry Pics
lol was a good night, coat kept me warm on the 45minute walk home in the rain :D
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Drinks Curry Drinks!!!!!!
yupyupyupyupyupyupyupyup
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just built a new 'puter
what motherboard did you get? is the 64bit chip working well with 64bit windows?
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300zx TT kicks ass......
The silver zed is Gio's the 350 is john-boys both members here :D
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Leigh-On-Sea Curry and Drinks
yes
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the joy of shift work
on call here also, been quiet so far though :D
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Medal Of Honour Pacific Assault
upgrade drivers and upgrade to latest version of directx :)
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Winners of the 300zx.co.uk Christmas Raffle
even though i pointed it out first :P
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Merry Crimbo and a very Happy New Year everyone!
happy crimbo timmy, see you in the new year for farewll drinks
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Merry Christmas All
happy christmas mate
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Its Alive And Kicking Hard
woohoo good to have her back innit?
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As I'll Be Twatted Later.........
you were twatted many moons ago mate lol, merry christmas,
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circle centre
Finding the Center of a Circle from Three Points Date: 05/22/2000 at 10:18:50 From: Christian Furst Subject: Center point of circle I have the coordinates of three ordered points on a circle. I want to find a way to define the circle's center. The purpose is to find a way to draw the part of the circle that is connecting the three points (by knowing the center and radius). -------------------------------------------------------------------------------- Date: 05/22/2000 at 15:34:50 From: Doctor Rob Subject: Re: Center point of circle Thanks for writing to Ask Dr. Math, Christian. Let the equation of the circle be (x-h)^2 + (y-k)^2 = r^2, and substitute the three known points, getting 3 equations in 3 unknowns h, k, and r: (x1-h)^2 + (y1-k)^2 = r^2 (x2-h)^2 + (y2-k)^2 = r^2 (x3-h)^2 + (y3-k)^2 = r^2 which you can solve simultaneously. First subtract the third equation from the other two, thus eliminating r^2, h^2, and k^2. That will leave you with 2 simultaneous linear equations in h and k to solve. This you can do as long as the 3 points are not collinear. Then those values of h and k can be used in the first equation to find the radius: r = sqrt[(x1-h)^2 + (y1-k)^2]. Example: Suppose a circle passes through the points (4,1), (-3,7), and (5,-2). Then we know that: (h-4)^2 + (k-1)^2 = r^2 (h+3)^2 + (k-7)^2 = r^2 (h-5)^2 + (k+2)^2 = r^2 Subtracting the first from the other two, you get: (h+3)^2 - (h-4)^2 + (k-7)^2 - (k-1)^2 = 0, (h-5)^2 - (h-4)^2 + (k+2)^2 - (k-1)^2 = 0, h^2 + 6h + 9 - h^2 + 8h - 16 + k^2 - 14k + 49 - k^2 + 2k - 1 = 0 h^2 - 10h + 25 - h^2 + 8h - 16 + k^2 + 4k + 4 - k^2 + 2k - 1 = 0 14h - 12k + 41 = 0 -2h + 6k + 12 = 0 10h + 65 = 0 30h + 125 = 0 h = -13/2 k = -25/6 Then r = sqrt[(4+13/2)^2 + (1+25/6)^2] = sqrt[4930]/6 Thus the equation of the circle is: (x+13/2)^2 + (y+25/6)^2 = 4930/36 Understood?