If Formula Rossa was a Strata-coaster...
- Thread starter Temleh
- Start date
Smithy
Strata Poster
Using very crude logic;
Stealth is 205 feet tall, divided by 80 for mph, is 2.56.
Kingda Ka is 456 feet tall, divided by 128 for mph, is 3.56.
So going off of Kingda Ka's horrendously crude workings out above, you could reach 530 feet, going off of Stealth's horrendously crude workings out above, you could reach 381 feet. Which suggests (again, loosely) that the taller the top hat goes, or the longer the vertical period, the less speed that is required to do so.
I'd say 500 feet at least would be expected, which is as much as a logical guess as the drivel above.
Stealth is 205 feet tall, divided by 80 for mph, is 2.56.
Kingda Ka is 456 feet tall, divided by 128 for mph, is 3.56.
So going off of Kingda Ka's horrendously crude workings out above, you could reach 530 feet, going off of Stealth's horrendously crude workings out above, you could reach 381 feet. Which suggests (again, loosely) that the taller the top hat goes, or the longer the vertical period, the less speed that is required to do so.
I'd say 500 feet at least would be expected, which is as much as a logical guess as the drivel above.
Keep in mind that air resistance is proportional to the square of the velocity. In short, for every kph you accelerate to, the harder it would be to accelerate to go one additional kph faster. The faster you go, the more the wind will slow you down. Thus, for every metre you add to the height of the coaster, the more force you have to put into the train.
Kingda Ka is ~13 kph faster than TTD, and rougly 10 metres taller. Formula Rossa is 36 kph faster than Kingda Ka, but I doubt that extra initial speed would help it go much more than 10-15 additional metres taller, even though it's almost 3 times as much difference in speed as between Ka and TTD. At such high speeds, air resistance gets pretty formidable, and every kph makes it increasingly worse.
Kingda Ka is ~13 kph faster than TTD, and rougly 10 metres taller. Formula Rossa is 36 kph faster than Kingda Ka, but I doubt that extra initial speed would help it go much more than 10-15 additional metres taller, even though it's almost 3 times as much difference in speed as between Ka and TTD. At such high speeds, air resistance gets pretty formidable, and every kph makes it increasingly worse.
Expanding Smithy's logic. And to demonstrate Poke's point as he finished his post before mine and made a similar conclusion.
If the kinetic energy of the train at the bottom of the hill is all converted into potential energy at the top, this gives you the 'ideal' case (ie, no friction, no air resistance, the ride comes to rest completely on the top of the top hat). If you make the assumption that the trains all weight the same and that they all crest the top hats at the same speed (a suitable enough assumption for the time being), you can treat all of the the lost energy, plus the kinetic energy at the crest, as a 'loss' term.
So KE(at the bottom)=PE(at the top) + loss
The neat(ish) thing with that is that it means you can get rid of the mass terms.
KE=0.5*m*v^2
PE=m*g*h
In other words:
0.5*m*v^2 = m*g*h + loss
OR
0.5*v^2=g*h + loss/m
That's convenient, because we can sort of roll the mass all into the generic loss term. If we then work out the KE at the bottom, and the PE the train would have at the top, we can work out this loss term. Doing this, you get numbers which look like:
Stealth:
Speed = 35.8 [m/s]
Height* = 62.5 [m]
loss=KE-PE=639.5-613.3=26.2
TTD:
Speed = 53.6 [m/s]
Height* = 128 [m]
loss=KE-PE=1438.9-1255.8=183.0
KK:
Speed = 57.2 [m/s]
Height* = 139.0 [m]
loss=KE-PE=1637.1-1363.5=273.6
*this is also ignoring the fact that they don't all start from ground level, but if we assume the launch track for each of them is at the same height then this is sort of cancelled out.
This is pretty clear that the faster these trains go the higher the losses are. If you were to assume (and this is a pretty massive assumption) that the loss terms are varying quadratically with speed - which is nice since air resistance (probably the single biggest factor slowing the trains down) is a function of the velocity squared, you find that the loss terms for Formula Rossa:
Speed = 66.7 [m/s]
loss=562.5 (found using the power-best fit line in Excel so it looks sort of right)
Plugging that back in, gives you a height of 169.1m, or 554.8ft.
There's a lot of big assumptions in there, and it's really not achieved much more than Smithy's much shorter estimation, but a second method that puts the numbers roughly in the same ball park serves as a pretty good validation that something around 500-530ft would probably be possible with Formula Rossa's launch.
[/overthetop]
EDIT: For clarification. Smithy's eyeball-estimate was pretty good, and Poke makes the most important point that keeping going faster and faster only really gives you more and more to fight against and so doesn't have as a dramatic effect as you might imagine.
If the kinetic energy of the train at the bottom of the hill is all converted into potential energy at the top, this gives you the 'ideal' case (ie, no friction, no air resistance, the ride comes to rest completely on the top of the top hat). If you make the assumption that the trains all weight the same and that they all crest the top hats at the same speed (a suitable enough assumption for the time being), you can treat all of the the lost energy, plus the kinetic energy at the crest, as a 'loss' term.
So KE(at the bottom)=PE(at the top) + loss
The neat(ish) thing with that is that it means you can get rid of the mass terms.
KE=0.5*m*v^2
PE=m*g*h
In other words:
0.5*m*v^2 = m*g*h + loss
OR
0.5*v^2=g*h + loss/m
That's convenient, because we can sort of roll the mass all into the generic loss term. If we then work out the KE at the bottom, and the PE the train would have at the top, we can work out this loss term. Doing this, you get numbers which look like:
Stealth:
Speed = 35.8 [m/s]
Height* = 62.5 [m]
loss=KE-PE=639.5-613.3=26.2
TTD:
Speed = 53.6 [m/s]
Height* = 128 [m]
loss=KE-PE=1438.9-1255.8=183.0
KK:
Speed = 57.2 [m/s]
Height* = 139.0 [m]
loss=KE-PE=1637.1-1363.5=273.6
*this is also ignoring the fact that they don't all start from ground level, but if we assume the launch track for each of them is at the same height then this is sort of cancelled out.
This is pretty clear that the faster these trains go the higher the losses are. If you were to assume (and this is a pretty massive assumption) that the loss terms are varying quadratically with speed - which is nice since air resistance (probably the single biggest factor slowing the trains down) is a function of the velocity squared, you find that the loss terms for Formula Rossa:
Speed = 66.7 [m/s]
loss=562.5 (found using the power-best fit line in Excel so it looks sort of right)
Plugging that back in, gives you a height of 169.1m, or 554.8ft.
There's a lot of big assumptions in there, and it's really not achieved much more than Smithy's much shorter estimation, but a second method that puts the numbers roughly in the same ball park serves as a pretty good validation that something around 500-530ft would probably be possible with Formula Rossa's launch.
[/overthetop]
EDIT: For clarification. Smithy's eyeball-estimate was pretty good, and Poke makes the most important point that keeping going faster and faster only really gives you more and more to fight against and so doesn't have as a dramatic effect as you might imagine.
Jordanovichy
Credit Whore 2016
Not often you get a physics lesson on CoasterForce 
Taxi
Mega Poster
If Formula Rossa were 'taller' it's height would stretch well beyond the 400-foot range.
With a velocity of 66.7 m/s, the roller coaster could theoretically reach a height of 226.8 meters, or 744.2 feet when starting from ground level. However, that is omitting all friction & air-resistance (the real figure would be less). And this 226.6 m figure is the apex, so it would have to be shorter to include the full top hat. So this model is basically a S:EFK/TTII clone
(omitted all the boring math and just gave a straight answer)
Source: big physics nerd
With a velocity of 66.7 m/s, the roller coaster could theoretically reach a height of 226.8 meters, or 744.2 feet when starting from ground level. However, that is omitting all friction & air-resistance (the real figure would be less). And this 226.6 m figure is the apex, so it would have to be shorter to include the full top hat. So this model is basically a S:EFK/TTII clone
(omitted all the boring math and just gave a straight answer)
Source: big physics nerd