On 26.06.13 10:26, Jon Elson wrote: > Erik Christiansen wrote: > > On 26.06.13 01:30, Gregg Eshelman wrote: > > > >> If you're near a coast or large/deep lake you might find a deal on a > >> used electric downrigger. Load it up with 300 pound test line and > >> there's your mechanicals, just need to figure out how to brake it to a > >> stop at the right moment without breaking things. ;-) > >> > > > > Careful; 200 pound chandelier x 1.5g deceleration = 300 pound force. > > > Oops - error. A 1.5 G stop of a falling 200 Lb object requires 200 + > 300 = 500 Lbs > tension on the cable.
No, that is in error, I believe, and the original statement stands. A 1.5g deceleration is one where the decelerating mass experiences 1.5g of acceleration forces. Not 2.5 g. To maintain a downward 1.5g deceleration in a 1g gravity well, the rate of change of velocity will only be 1/3 of that to produce a 1.5g effect in the horizontal plane. At all times, the force on the cable is 300 pounds while the mass is experiencing 1.5g, even if 2/3 of that is the acceleration due to gravity. Erik -- "You are serving roast hare today, I read outside", said the official from trade inspection. "Pure roast hare?" "With some horse", confessed the chef. "In what ratio?" "50 - 50: One hare, one horse." ------------------------------------------------------------------------------ This SF.net email is sponsored by Windows: Build for Windows Store. http://p.sf.net/sfu/windows-dev2dev _______________________________________________ Emc-users mailing list [email protected] https://lists.sourceforge.net/lists/listinfo/emc-users
