QUOTE: Originally posted by mudchicken KenEO: Not trying to "blow smoke" and can understand the "ballpark" reference 7 or so posts back...What is frightening is the people that take stuff in this forum and others and think this stuff is gospel. If I had a nickel for every time I corrected the work of a non-railroad licensed engineer or surveyor that has encountered a railroad, I could retire and live off the royalties. Some of the assumptions, misconceptions and mistakes should be published in Ripley's. (Some still have us in stitches, they are that bizzare! Others make me concerned about my chosen profession and the public's safety)...Like Ed, just trying to keep things "somewhere in the middle".... We all get to learn a little and have a little fun here.
QUOTE: Originally posted by Mookie Eric - I read that and actually understood it! All except the # 1 Market - ? Jen
She who has no signature! cinscocom-tmw
QUOTE: Originally posted by edblysard I dont think the height of the person would make much difference. It would all depend on.. ouch= toe+ rail+rocksX2 if someone is watching divided by density of skull. But with some of the guys I work with, it wouldnt matter how deep the dent went. Nothing in there to hurt in the first place.[:D] Stay Frosty, EdQUOTE: Originally posted by kenneo Gives "trip" a whole new meaning. Ed, does this now mean that a 5 foot person falling victim to your foumula would ge a dented forhead and a massive non-migrain headache? How would that alter your formula? (big wide cheesy grin!)
QUOTE: Originally posted by kenneo Gives "trip" a whole new meaning. Ed, does this now mean that a 5 foot person falling victim to your foumula would ge a dented forhead and a massive non-migrain headache? How would that alter your formula? (big wide cheesy grin!)
23 17 46 11
QUOTE: Originally posted by mudchicken Kenneo- carefull, speed is a function of radius/degree of cv. and elevation, switches for the the most part are flat...plus unbalance (in relation to centripital force) SPEED= SQRT((E+3)/0.0007D) E=0, (flat) D=degree of curvature, 3" maximum unbalance (freight railroads use 2 or less) Mookie's closed book/open mind pop quiz to follow....... The proper tem for "Y" switch is equilateral turnout, and they are a relatively minority item whose geometry precludes it from many locations.
Carl
Railroader Emeritus (practiced railroading for 46 years--and in 2010 I finally got it right!)
CAACSCOCOM--I don't want to behave improperly, so I just won't behave at all. (SM)
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