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Pick the right venue

I’m a fan of the scale where my brother swims. It reported me, before we headed up Katahdin, as less than one cat over marathon weight. I think that’s probably optimistic, but not wildly inaccurate.

The scale in the fieldhouse, however, says I’m nearly half a cat under weight, which strikes me as flat-out absurd. Sure, I’ve been eating erratically lately (every now and then a schedule is too tight to allow lunch,) and probably less, overall, than before. But not that much.

Posted by pjm on September 13, 2005 8:59 AM |

Comments

The obvious solution here is that your mass is constant and the cat’s mass doubled. m + C1 = P1 m - 0.5(C2) = P1

2*C1 = C2

Posted by: Anonymous | September 13, 2005 2:40 PM

You’ve got a sign issue: if

m + C1 == m - 0.5(C2) // Follows from postulates 1 and 2

then

C1 == -0.5(C2) // Subtraction - “m”s cancel.

since m is also a constant. Your solution requires the cat to have negative mass, which, while sometimes suggested by observation, is seldom the case when he’s sitting on your chest asking for breakfast.

Posted by: pjm | September 13, 2005 4:17 PM

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