The alarm went off a 5:30 this morning, so we could leave home at 6:30 for our flight at 9:00. We were flying Delta out of the A concourse this time (usually it's B) which, we found out, has an excellent kids play area (even tho it's much smaller and generally has much less stuff than B). Plastic houses and slides and bridges and everything. Plus now I think it has Amelia's sippy cup, because that's the last place we remember seeing it.

Amelia was very kind to us, and slept for most of the two hour flight.

It's much warmer in Jacksonville, Florida than it is in Cincinnati, Ohio right now.

We're picked up at the airport by Ruth Anne's father, and taken to Jen and Jack's palatial estate. The only thing that keeps me from being totally jealous is that it's on a suburban street fulls of dozens of identical palatial estate.

They have a DSL modem, but it took a while to figure out how to make it work for me. PPPoE? What is this, the dark ages? A bunch of trial and error ensued, hunting around on Jen's Windows box for the settings I need on my Macintosh. Turns out all I needed was PPPoE, the name and password, and check the "connect automatically when needed" box. Poof! I'm back online.

My father in law told me about this experiment where you take two heavy metal balls and suspend them from long wires. Allegedly, one can measure the distance between the wires at the top and bottom with a tape measure, and see a difference which can be attributed to the gravitational pull of the two balls on each other. We both assumed that this was poppycock, and proceeded to try to run some numbers.

Picking some numbers

  • L : length of wire : 10 meters
  • m : weight of balls : 10 Kg
  • r : distance between centers of balls 0.1 meter
  • D : minimum measurable displacement .001 meter
Some rough constants :
  • g : 10 m/s^s
  • M : mass of earth : 6 x 10^24 Kg
  • R : radius of earth : 6378 Km

The displacement could be measured like this : measure the horizontal displacement D of the ball and the length L of the wire. the angle theta would be arcsin(D/L). the force necessary for that displacement would be the mass m of the ball times the sin of the angle or m * sin(arcsin(D/L)) or m*D/L. The ratio T1 of this force to the force pulling down is (mD/L)/(mg) or D/gL. For our numbers above, this gives a ratio of .001/10*10 or 10^-5. Thus, the ratio of the forces must be at least 10^5 to have a measurable displacement.

The attraction between two bodies is given by Gm1m2/r2 Thus, the force between balls is Gmm/rr and the force between the earth and the ball is GmM/RR. The ratio of these forces is (m/rr)/(M/RR) or mRR/Mrr. For our numbers above this makes the ratio of the forces (10*6378000*6378000)/(6x10^24*0.1*0.1) or (10*4x10^13)/(6x10^24*10^-2) or 4x10^14/6x10^22 or 2/3 * 10^-8. Thus the actual ratio of forces is around 10^8, which is three orders of magnitude too small.

Please find holes in my reasoning and/or calculations.