Physics question - Slingshot Ride but not strapped in
June 30, 2021 7:25 AM
Ok, so let's say you got on the slingshot ride at the carnival, but all the safety harnesses were removed. And then they launched you. Based on publicly available information about the velocity and height of these rides, how far do you think the riders would be launched into the air before they started falling?
As a quick first answer, you'd go as far as normal more or less.
First, these rides have no lateral force, it's all up and down, so to a first approximation you'd just fall straight down, not get launched halfway across the midway into a corndog stand. (In the real world, there will be wind and things won't be perfectly up and down so you might travel a few feet in any horizontal direction, but not enough to be important.)
Second, let's consider the vertical dimension. There is an upward force imparted to you and the cabin by the bungees, then gravity pulls you down, with air resistance slowing things a little in each direction. Both the upward force and gravity are proportional to weight; if the cabin weighs 3 times as much as you, it'll have 3 times as much upward force and 3 times as much gravity. So with just these forces, you're going to go basically as high as the cabin does normally.
The bungees might provide a little downward force at the top of the arc, in which case you'd fly a few feet higher than the cabin since they would be providing an additional downward force on just the cabin. You could watch YouTube videos to check this out; do the cables remain slack or get taut at the top of the ride? If slack, no more force. If taut, then there's a little bit more force being applied.
As a first approximation, I'd guess the riders would travel higher than the cabin proportional to their relative weights. So if the cabin weighs 400 pounds and the riders 200 pounds (2/3 of the weight is cabin) and the cabin-and-riders normally travel 30 feet vertically once the lines become taut, I'd expect if separated the cabin to only go 20 feet (2/3 of the distance) and the riders to go 45 feet (3/2 of the distance). Note this is only the distance once the lines become taut, which is a small fraction of the total vertical distance.
So if a reverse bungee shoots you to a height of 200 feet vertically, if you weren't in your harness you might go 210 or 225 feet or something like that; it wouldn't be a Wile E. Coyote type situation. Which is not to diminish how unpleasant falling 225 feet unrestrained onto the tarmac might be.
posted by Superilla at 8:01 AM on June 30, 2021
First, these rides have no lateral force, it's all up and down, so to a first approximation you'd just fall straight down, not get launched halfway across the midway into a corndog stand. (In the real world, there will be wind and things won't be perfectly up and down so you might travel a few feet in any horizontal direction, but not enough to be important.)
Second, let's consider the vertical dimension. There is an upward force imparted to you and the cabin by the bungees, then gravity pulls you down, with air resistance slowing things a little in each direction. Both the upward force and gravity are proportional to weight; if the cabin weighs 3 times as much as you, it'll have 3 times as much upward force and 3 times as much gravity. So with just these forces, you're going to go basically as high as the cabin does normally.
The bungees might provide a little downward force at the top of the arc, in which case you'd fly a few feet higher than the cabin since they would be providing an additional downward force on just the cabin. You could watch YouTube videos to check this out; do the cables remain slack or get taut at the top of the ride? If slack, no more force. If taut, then there's a little bit more force being applied.
As a first approximation, I'd guess the riders would travel higher than the cabin proportional to their relative weights. So if the cabin weighs 400 pounds and the riders 200 pounds (2/3 of the weight is cabin) and the cabin-and-riders normally travel 30 feet vertically once the lines become taut, I'd expect if separated the cabin to only go 20 feet (2/3 of the distance) and the riders to go 45 feet (3/2 of the distance). Note this is only the distance once the lines become taut, which is a small fraction of the total vertical distance.
So if a reverse bungee shoots you to a height of 200 feet vertically, if you weren't in your harness you might go 210 or 225 feet or something like that; it wouldn't be a Wile E. Coyote type situation. Which is not to diminish how unpleasant falling 225 feet unrestrained onto the tarmac might be.
posted by Superilla at 8:01 AM on June 30, 2021
I'll let someone better informed than I am show the physics calculations, but you may find the death of Jérôme Charron of some tragic interest. As well as this question for the Guardian, which says 30m seems to be the limit for parachuting.
posted by Nonsteroidal Anti-Inflammatory Drug at 8:02 AM on June 30, 2021
posted by Nonsteroidal Anti-Inflammatory Drug at 8:02 AM on June 30, 2021
I believe they are "falling" the instant the forces imposed on them by the ride go to zero
That would only be the case if there was no restraining effect of the rider at the top of the rides arc. So you'd also need to be zero force from the ride, *plus* also no vertical speed/momentum. That would need to dissipate first before you fell. From the videos I have watched there is a slowing effect on the cradle of the ride as it gets to the top but it is close to finishing rising anyway. (based on the first few minutes of this: https://www.youtube.com/watch?v=PzI7cvWUAA8)
Gut feeling is if you were entirely unrestrained you'd only maybe come out of your seat a bit. The cradle is slowing pretty well towards the end of its path so you'd be maybe 1-2 feet above the seat at the absolute most and then you'd just follow it down. Safety standards would suggest that the restraints aren't stopping you from flying out and the design is such that you're basically being held in place by the forces and the restraints are belt and braces overkill.
My feeling is you'd still be within the confines of the cage as you got to peak height, but the cage would accelerate away from you due to the tether effect of the straps on the cage. So you'd maybe be 6 feet above it pretty quickly.
posted by Brockles at 8:02 AM on June 30, 2021
That would only be the case if there was no restraining effect of the rider at the top of the rides arc. So you'd also need to be zero force from the ride, *plus* also no vertical speed/momentum. That would need to dissipate first before you fell. From the videos I have watched there is a slowing effect on the cradle of the ride as it gets to the top but it is close to finishing rising anyway. (based on the first few minutes of this: https://www.youtube.com/watch?v=PzI7cvWUAA8)
Gut feeling is if you were entirely unrestrained you'd only maybe come out of your seat a bit. The cradle is slowing pretty well towards the end of its path so you'd be maybe 1-2 feet above the seat at the absolute most and then you'd just follow it down. Safety standards would suggest that the restraints aren't stopping you from flying out and the design is such that you're basically being held in place by the forces and the restraints are belt and braces overkill.
My feeling is you'd still be within the confines of the cage as you got to peak height, but the cage would accelerate away from you due to the tether effect of the straps on the cage. So you'd maybe be 6 feet above it pretty quickly.
posted by Brockles at 8:02 AM on June 30, 2021
Oh, and if my answer seems unintuitive based on your experience of a slingshot, note two key differences: 1, in the slingshot, you're going horizontally, not directly against gravity. 2, in a slingshot, you're shooting something very small relative to the weight of the sling, where I'd assume the cabins of these rides to weigh as little as possible given that every extra pound means buying stronger bungees. So most of the weight in the cabin and riders is probably actually in the riders.
posted by Superilla at 8:09 AM on June 30, 2021
posted by Superilla at 8:09 AM on June 30, 2021
Fun question! Note that I had to look up what a slingshot ride was, so take this with an appropriate grain of salt. Some random websites suggest slingshot rides reach a maximum velocity between 60 and 100 miles per hour depending on the ride. Presumably, if you're mostly going straight up, this happens when the line becomes slack and the chair is no longer accelerating you. After that you're moving vertically and subject to only the force of gravity.
For an object traveling straight up with gravitational acceleration g, the heigh x will be given by x=x_0+v_0*t + 1/2 g t^2, where x_0 is the height where the line became slack, v_0 is the velocity when that happened, g is 9.8 m/s^2, and t is time in seconds. One way to solve it is set the derivative dx/dt =0, which happens when you stop rising and begin to fall, solve for t, then plug it back in to the equation for x.
For 100 miles/hr, you'd reach a maximum height in 4.6 seconds, 103 meters above the position where the lines went slack. For 60 miles/hr, your read a maximum in 2.7 seconds, 37 meters higher than where the line went slack. (Unless I screwed up the arithmetic.) Where the lines go slack - or where velocity is maximum - compared the height of the tower could be guessed from videos or by someone who knows more about bungee cords than me. If you have significant horizontal motion, it's also straightforward to solve, but you'd need some estimate of how that compares to your vertical speed.
posted by eotvos at 8:37 AM on June 30, 2021
For an object traveling straight up with gravitational acceleration g, the heigh x will be given by x=x_0+v_0*t + 1/2 g t^2, where x_0 is the height where the line became slack, v_0 is the velocity when that happened, g is 9.8 m/s^2, and t is time in seconds. One way to solve it is set the derivative dx/dt =0, which happens when you stop rising and begin to fall, solve for t, then plug it back in to the equation for x.
For 100 miles/hr, you'd reach a maximum height in 4.6 seconds, 103 meters above the position where the lines went slack. For 60 miles/hr, your read a maximum in 2.7 seconds, 37 meters higher than where the line went slack. (Unless I screwed up the arithmetic.) Where the lines go slack - or where velocity is maximum - compared the height of the tower could be guessed from videos or by someone who knows more about bungee cords than me. If you have significant horizontal motion, it's also straightforward to solve, but you'd need some estimate of how that compares to your vertical speed.
posted by eotvos at 8:37 AM on June 30, 2021
Also, if you wanted to do this calculation for a specific ride that you have video of, physlets tracker might be a useful tool. You'll need some way to set a scale in physical units based on some object in the video and the assumption that the video frames aren't sped up or slowed down. It does lots of fancy stuff, but can also just be used to make a plot or table of velocity and position vs. time.
posted by eotvos at 8:48 AM on June 30, 2021
posted by eotvos at 8:48 AM on June 30, 2021
Also, also, and with apologies for thread-squatting, note that my answer doesn't include any air resistance. 100 miles/hr is pretty close to terminal velocity for a horizontal human, so it will be important in the beginning. It'll reduce the final height by a bit. (Figuring out by how much requires looking up drag coefficients for humans in appropriate positions. Which I'm sure some military research org has published.)
posted by eotvos at 8:56 AM on June 30, 2021
posted by eotvos at 8:56 AM on June 30, 2021
*plus* also no vertical speed/momentum
I think what achrise is getting at is that the moment you have no force acting on you but gravity (or an equivalent statement, your only acceleration is toward the ground due to gravity), even if you still have a large amount of velocity in any direction (including up), you are by definition in freefall, and could therefore be said to be falling. That doesn't help answer the question "at what point will the person start falling *downwards*," (which is to me the more likely interpretation of the question) but it does answer the very similar question "at what point will the person start feeling like they are falling."
posted by solotoro at 9:03 AM on June 30, 2021
I think what achrise is getting at is that the moment you have no force acting on you but gravity (or an equivalent statement, your only acceleration is toward the ground due to gravity), even if you still have a large amount of velocity in any direction (including up), you are by definition in freefall, and could therefore be said to be falling. That doesn't help answer the question "at what point will the person start falling *downwards*," (which is to me the more likely interpretation of the question) but it does answer the very similar question "at what point will the person start feeling like they are falling."
posted by solotoro at 9:03 AM on June 30, 2021
You'd come off well before the ride sped up fully. If released at full speed, well, didn't you ever play gorillas?
Read about the Verrukt water slide at Schlitterbahn. This question is not entirely theoretical. Content Warning: I read the MeFi link when it was posted and still think about the tragedy.
previously, also
posted by theora55 at 11:32 AM on June 30, 2021
Read about the Verrukt water slide at Schlitterbahn. This question is not entirely theoretical. Content Warning: I read the MeFi link when it was posted and still think about the tragedy.
previously, also
posted by theora55 at 11:32 AM on June 30, 2021
This thread is closed to new comments.
posted by achrise at 7:49 AM on June 30, 2021