
Good thing the makers of the first rollercoaster hadn't read this - would have been a shame if they got dissuaded!
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Sadly, they never took off. Fortunately, they never took off.
ctdonath wrote:I'm just not quite seeing the reasoning on this one.
Maybe I need coffee.
I'll go get some coffee.
Maelin wrote:it's not at all clear why a fast train on a large loop might be okay while a slow train on a small loop wouldn't.
Pfhorrest wrote:As someone who is not easily offended, I don't really mind anything in this conversation.
Mighty Jalapeno wrote:It was the Renaissance. Everyone was Italian.
rhomboidal wrote:Whoever does manage to build one will need to include one of those restriction signs: "You Must Be At Least This Crazy/Brave/Just-Plain-Dumb" To Ride"
blueberry42 wrote:I think this comic is wrong.
Consider a circular loop with radius r, speed at bottom of loop v0, and speed at top of loop v1. Assume train of mass m is just coasting, i.e. no engines. By conservation of energy we have
(1/2)m v02 = (1/2)m v12 + mg(2r).
Multiply both sides by 2/(mr), yielding
v02 / r = v12 / r + 4 g.
Centripetal acceleration is velocity-squared over radius, so this means the centripetal acceleration at the top is 4 gees less than at the bottom. So if we enter at 5 gees centripetal acceleration the centripetal acceleration at the top is 1 gee, and hence the train stays on the track (barely). The gee-force felt at the bottom would be 6 gees (extra g from gravity), which is uncomfortable but perfectly survivable.
blueberry42 wrote:I think this comic is wrong.
Consider a circular loop with radius r, speed at bottom of loop v0, and speed at top of loop v1. Assume train of mass m is just coasting, i.e. no engines. By conservation of energy we have
(1/2)m v0^2 = (1/2)m v1^2 + mg(2r).
Multiply both sides by 2/(mr), yielding
v0^2 / r = v1^2 / r + 4 g.
Centripetal acceleration is velocity-squared over radius, so this means the centripetal acceleration at the top is 4 gees less than at the bottom. So if we enter at 5 gees centripetal acceleration the centripetal acceleration at the top is 1 gee, and hence the train stays on the track (barely). The gee-force felt at the bottom would be 6 gees (extra g from gravity), which is uncomfortable but perfectly survivable.
Angelastic wrote:I'm disappointed that he gives numbers for various possible, impossible, limiting, 'great' loops, but then the final answer is just 'medium-sized' 'something like this' and 'fast'. Actually, we don't even get any numbers for speeds except for '35% faster'. I'd rather know a bit more about looping trains and less about where Joe Biden eats his sandwiches.
At the train’s top speed, that curve would create about two gees of acceleration. This might be survivable (for the passengers, at least, if not the train), but it would certainly not be comfortable.
Sabbede wrote:I'd just like to point out that the reason Joe Biden's commute was so dull has less to do with the length of the trip then it does with the fact that Joe is pretty much the only person in the country who rides AmTrak.
blueberry42 wrote:I think this comic is wrong.
Consider a circular loop with radius r, speed at bottom of loop v0, and speed at top of loop v1. Assume train of mass m is just coasting, i.e. no engines. By conservation of energy we have
(1/2)m v0^2 = (1/2)m v1^2 + mg(2r).
Multiply both sides by 2/(mr), yielding
v0^2 / r = v1^2 / r + 4 g.
Centripetal acceleration is velocity-squared over radius, so this means the centripetal acceleration at the top is 4 gees less than at the bottom. So if we enter at 5 gees centripetal acceleration the centripetal acceleration at the top is 1 gee, and hence the train stays on the track (barely). The gee-force felt at the bottom would be 6 gees (extra g from gravity), which is uncomfortable but perfectly survivable.
blueberry42 wrote:I think this comic is wrong.
Consider a circular loop with radius r, speed at bottom of loop v0, and speed at top of loop v1. Assume train of mass m is just coasting, i.e. no engines. By conservation of energy we have
(1/2)m v0^2 = (1/2)m v1^2 + mg(2r).
Multiply both sides by 2/(mr), yielding
v0^2 / r = v1^2 / r + 4 g.
Centripetal acceleration is velocity-squared over radius, so this means the centripetal acceleration at the top is 4 gees less than at the bottom. So if we enter at 5 gees centripetal acceleration the centripetal acceleration at the top is 1 gee, and hence the train stays on the track (barely). The gee-force felt at the bottom would be 6 gees (extra g from gravity), which is uncomfortable but perfectly survivable.
Most high-speed trains are limited to vertical curves with radii no shorter than 20 kilometers.
The reason for those limits isn't that trains aren't bendy enough. It's how fast they're going.
Patrik3 wrote: Someday, I'll get around to working out the formula for a 'perfect' loop - that is, one where the train enters at speed V and where the rider experiences a constant acceleration G + A throughout. No spoilers, please!
tibfulv wrote:As a Norwegian, I have to say that this was not funny.
vvn wrote:tibfulv wrote:As a Norwegian, I have to say that this was not funny.
As a Norwegian-American with great pride in my Norwegian ancestry, I have to say it was funny. And, I can't see any way to take it in an insulting manner. (I also think it's not doable, but not for the reasons Randall has expressed.)
tibfulv wrote:Sorry, bad joke. Just trying to parody the people who take offense at anything.
Adam H wrote:tibfulv wrote:Sorry, bad joke. Just trying to parody the people who take offense at anything.
As one of those people that take offense at anything, I take serious offense.
aadams wrote:I am a very nice whatever it is I am.
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