OK, I won't argue any more. Just so long as 1+1=2, payday's on Friday, the beer is cold (sorry, Ian), and somebody finds out why mosquitoes were included on the Ark.
We had a late thaw this year, went out this morning to do some work and was greeted by a swarm of mosquitoes. They were big enough that I could see the insignia on their wings and bodies, pretty bad when you can count the kill marks! Had six feeding on one arm, and 20 more lined up making reservations. Got so bad, the swallows were hiding from them all; crows were carrying shotguns in defense, and the eagles thought they'd died and gone to heaven and thought they were feeding on sparrows. Have I told you about my cousin, Jasper? Took his bull, Fred, into town the other day, to get it shod. Trouble was, Farmer Brown was moving his herd of cows to market. Fred got excited and started chasing the truck, which he caught about 5 miles down the road. RCMP came along, gave Jasper a ticket for unsafe operation of a bull, Farmer Brown a ticket for running a truck of ill repute, and Fred a ticket for speeding. Nothing for the girls, however, just Fred. Stay tuned for more Cariboo humor. Pat -----Original Message----- From: Daniel Carrera [mailto:[EMAIL PROTECTED] Sent: 2008/05/22 05:29 To: [email protected] Subject: Re: [social] Definition of metre in other planets Pat McBride wrote: > Read Science Fiction, Daniel? You're right on, but the derivation would > probably rattle a few brains, including mine. Haven't done any calculus for > going on nigh 35 years, been better off without it. No calculus, it's actually just basic algebra. I didn't include it before because I didn't think anybody would care, but here it is: Simplifying assumption: Assume the planet is uniform and homogeneous except possibly in the radial direction. Acceleration due to gravity is: g = GM/R^2 Where G is the gravitational constant, M is the mass of the planetary body, and R is its radius. M = rho*(4/3)*pi*R^3 Thus: g = G*rho*(4/3)*pi*R Where rho is the density of the earth. The period of a pendulum is: T = 2*pi*sqrt(L/g) Where L is the length of the pendulum. Let L be the length that corresponds to a 2-second pendulum. This is one of the two definitions of a metre that we wish to compare: L = g * (1s/pi)^2 L = G*rho*(4/3)*pi*R * (1s/pi)^2 L = G*rho*(4/3)* R * 1s^2 / pi Now consider the alternate definition of a metre (L') which is defined as 10 millionth of the length of a meridian measured from the equator of the planet to the pole: L' = 10^(-7) * (1/4) * (2*pi*R) 2*pi*R being just the circumference of the planet. We are interested in how L and L' compare. So we will investigate their ratio: L/L' = (G*rho*(4/3)* R * 1s^2/pi) / ( 10^(-7)*(1/4)*(2*pi*R) ) L/L' = (10^7*G*rho*(8/3)*1s^2) / (pi^2) Notice that this ratio does not include an R. So this ratio is independent of the radius of the planet. The only term that is not a constant is (rho) which is the density. Conclusion: The ratio between the two definitions of a metre depend only on the planet's density. Cheers, Daniel. --------------------------------------------------------------------- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] No virus found in this incoming message. Checked by AVG. Version: 8.0.100 / Virus Database: 269.24.0/1459 - Release Date: 08/05/21 17:34 --------------------------------------------------------------------- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED]
