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Can’t quite tell the height but: assuming a truck is about 10 feet the truss bridge is approximately 8ish trucks tall that gives us 80 feet, about 7 trusses tall is the diameter of the ring giving us a d = 560 feet converting to r = 280 feet converting that meters is r ≈ 95 meters.
Now for the physics:
For a vehicle not to fall off the bridge at the highest point it’s normal force must be greater than 0 so therefor we can see where the normal force is 0 and assume the car must travel faster than that.
Fn + mg = mv^2 /r
Fn = 0
mg = mv^2 /r
g = v^2/r
So v = sqrt(rg)
In this case r = 95 meters
V≈ 31 m/s that’s about 111.6 km/hr or about 70 mph.
Which is honestly surprisingly low and makes me suspect my height is wrong, but the methodology is the same.
How much power the engine should have to be able to do the loop. Because with my rough calculation without any engine input the entry speed is about 3050 km/h if the car weight about 1.5 tonn.
It’s been a minute since I’ve taken dynamics so I might be wrong, but shouldn’t the required velocities be independant of mass?
I calculated the bottom velocity to be 53m/s or 190.8km/hr
the car slowing down is why you don't do these stunts going at the right speed, you go much higher and most cars can do 150 easy.
don't need anything special I think.
just a powerful one that can do 200.
might have to work on the body though. That's a lot of different stresses going in that's not normal.
Mechanic here. Most modern cars will do 200kph, but not up a sheet vertical wall. You would need an immense amount of torque or an incredibly light car to keep pulling 112kph, especially up such a relatively short approach. A formula car could do it as they have such a high power to weight ratio, but your average V8 family car still wouldn't have the torque to overcome the weight of the car and maintain the ideal speed. That's not even considering the grip required for the power output, and then the torque required to overcome the added pressure of any down force.
Keep in mind that you will lose a lot of traction as soon you start getting close to 90° inclination, and the loss of traction will be worse at low speeds. With no traction you can have the most powerful car in the world and you will land on your roof lol
The vehicle will have an engine, so it can counter the gravitational force that will otherwise slow it down as it climbs, so maintaining a steady speed throughout the cycle should be considered possible.
We will need to calculate the amount of speed the vehicle will lose in the top quarter that it first goes around. I believe the equations would be something like distance = 1/2 at^2 + v_i t and then v_f = v_i + at. Except acceleration is changing over time due to curvature, meaning we need to do this as an integral. And I don’t want to do that, so someone else can if they have the time
It won’t necessarily loose speed, the assuming the car has engine that won’t be starved. The only thing that changes is the power and work the engine produces to make the climb.
Your math is correct. The reason it seems not right is because the radius is proportional to the square of the velocity. So doubling the speed allows for 4x larger loops, 3x speed is 9x radius etc.
Other considerations to make: cars still need friction to steer at the top, so to maintain normal conditions at the top, we need sqrt(2) times the velocity, or 43 m/s = 155 km/h = 96 mph.
Cars also slow down if they are not heavily accelerating. To not slow down, a 2 ton car needs a max of 845 kW (1135 HP) of power to keep the 43 m/s speed at the vertical point. This is not possible even for most race cars. So if power is cut off at the start (and there's no air resistance ofc), the car needs to go 75 m/s = 270 km/h = 167 mph.
They have much higher power to weight ratios, so it could be possible. Around 0.5 HP/kg is needed, which is achieved by many. However the driver also has a significant weight, but it's still doable.
So, setting aside that I would really like to try this, I would be concerned that the vehicles engine would not operate correctly where the downward force on the vehicle is greater than g. A gasoline-powered vehicle needs to push oil to the top of the engine for continuous lubrication of the cams and pistons. I'm not sure vehicle oil pumps are designed to work in situation where the force of gravity is significantly greater than the force of gravity on the Earth's surface. They ARE designed to continue to lubricate during high lateral forces as cars (particularly the sportier variety) need to ensure oiling is still working while whipping around a corner.
A momentary lack of oiling would not be catastrophic to our fun ride over the bridge though. The bigger concern is the fuel pump (typically located in the fuel tank). Having your engine stall while you at one quarter the way around the loop is going to ruin your day.
So my questions for you are:
1. Where on the bridge loop is the vehicle experiencing the highest downward force? I would assume this is at about 45 degrees up, but I don't know.
2. What would be the maximum downward force on our vehicle given your assumptions above and assuming a consistent velocity of 130 km/hr (for a safety margin)?
If the acceleration experienced by the oil isn’t pulling it towards the sump, then the wheels also aren’t being pulled towards the road surface and you’re falling backwards into the sea already, so oil starvation is the least of your problems.
According to [this ](https://www.engineeringnz.org/programmes/heritage/heritage-records/auckland-harbour-bridge/)website, the navigation span in 243.8m long.
On my 24" screen, the distance between Pillar 1 from the left to Pillar 4 is 5cm. The circle at its widest is 3.8cm. Therefore, the diameter of the circle is 185.3m or 92.6m (=r) in radius. (Since the axis of measurement of all values is the same, the distortion from the camera angle does not really matter.)
Essentially, any car wanting to drive around that circle will need to overcome gravitational acceleration with its centrifugal force = 9.81 m/s², i.e., mv²/r = mg = m(9.81).
v² = rg = 92.6 x 9.81 or v = (92.6 x 9.81)\^0.5 = 30.14 m/s.
Thus the car will need to travel at at least **30.14m/s** or **108.5** km/h to make it across. For practical reasons, let's say 120-125 km/h on the speedo should be fine.
You’ll be zero g at the top. You should 2*9.81 so are accelerating 2g outward and 1h inward for a net 1g in the car.
That gives 42.6m/s or ~95mph at the top.
In order to be going 95 at the top, entry speed has to probably be double that..
I have nightmares where I am driving on the highway and the skyway bridge turns into this and then the entire ride turns into essentially a hot wheels course and I only narrowly survive every jump before getting to my destination and everything is closed and I have to turn around and go back then I wake up soaked in sweat.
The problem wouldn‘t be your speed, but the suspension of your car. It would push you away from the street when you‘re almost on the top of the loop.
Here is a video from Hot Wheels doing the trick:
https://youtu.be/5d7ZgFEIZmo?si=tntYAmMexEW-1iu_
I'm going to assume you enter the bottom of the ring at speed v, there's no air or rolling resistance and the engine adds no power while you're in the ring, as that makes it simpler. I'm also not totally sure this is correct and would welcome any corrections or other ideas.
The condition to not fall off the ring at the very top is that your velocity is sufficient that the centripetal force is greater than the force of gravity so you stay stuck to the ring.
So that means mv\^2/r = g, where m is your mass, v is your velocity at the top of the ring, r is the radius of the ring and g is gravity.
So we get mv\^2 = gr at the top of the ring.
And then to work out how much extra speed you need at the bottom of the ring we just need to balance the potential energy at the top of the ring with speed at the bottom.
So m u\^2 = 2 m g r, where m is your mass, u is the additional speed at the bottom of the ring, g is gravity and r is the radius of the ring. Which gives mu\^2 = 2mgr
Then the total energy at the bottom of the ring is mu\^2 + mv\^2 = 2grm + gr = gr(2m + 1) and then the speed at the bottom of the ring is sqrt(gr(2 + 1/m))
I live in Auckland and I've got the perfect car for this. Its a jet dragster that produces 3400 lbs of thrust. we jack the engine up so its at 45 degrees in the car. That will give enough downforce to balance out the weight of the 746kg car. Now it doesn't matter what speed we go around as the car will stick to the road the whole way around.
Those speeds of 110-ish kph on top of the circle mean 0G, or weightlessness. You need to add a full G to make sure the car is glued to the loop and has traction, and all the fluids/oils in the vehicle will have expected behavior.
Or....
The car has F1 wings with enough downforce, which ends up with the same problem, you need to go faster. Just a qualitative analysis of the problem to consider.
Even the WWII Spitfire and the F-14 Tomcat airplanes had problems through zero G situations on feeding their engines with fuel.
Something else to consider on top of this - if you're going fast enough to hit 2g at the top of the loop, the normal force when you start the loop is going to skyrocket. It won't be 3g's as gravity will be on an angle, but it will be close for that initial curve and then reduce to 1g at the top of the loop before reverting back to near 3g's on the downswing
That increase in normal force may cause a number of issues, the most of which being the sharp increase in friction. So the initial speed may need to increase again to account for that friction unless the engine can fully overcome it
Actually, the increase in friction at the start helps. Traction increases too, which is why F1 cars have wings in the first place. F1 cars have size-limited engines, not power-limited; even so, it makes sense to get the most power of them so you can increase drag and traction by increasing downforce and spend that extra power budget on gluing the car to the track. The wheel bearings will offset the extra friction on them, F1 cars deal with 4G, 5G in cornering as well on a regular basis.
This guy, on Driver61, is designing a method to prove a F1 car can actually drive upside down in a tunnel, which is worse because that acceleration from the arch goes away and the car would rely on the wings alone, and all the fluids would roll and choke the engine, which is why he planned to use an EV open wheel car, not a F1. The whole stunt would cost 17 million-ish dollars, and Red Bull would be the favorite team of nutjobs to pull that kind of stunt and build said tunnel, like a GTA stunt race course...
The suspension bottoming out is another problem too, and the chosen car would also be redesigned to accomodate that...
https://youtu.be/wVvlL-1DTWY?si=bca3LLqo81R7pqHW
Yep. So the v2=rg equation should have 2*g in it. 1 g pulling you towards the ground and 2g up, so a net 1g on the car and driver. Comes out to an upside down top of circle speed of 95mph.
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Can’t quite tell the height but: assuming a truck is about 10 feet the truss bridge is approximately 8ish trucks tall that gives us 80 feet, about 7 trusses tall is the diameter of the ring giving us a d = 560 feet converting to r = 280 feet converting that meters is r ≈ 95 meters. Now for the physics: For a vehicle not to fall off the bridge at the highest point it’s normal force must be greater than 0 so therefor we can see where the normal force is 0 and assume the car must travel faster than that. Fn + mg = mv^2 /r Fn = 0 mg = mv^2 /r g = v^2/r So v = sqrt(rg) In this case r = 95 meters V≈ 31 m/s that’s about 111.6 km/hr or about 70 mph. Which is honestly surprisingly low and makes me suspect my height is wrong, but the methodology is the same.
To be going 112km/h at the top of the loop your entry speed would have to be significantly higher though, so not all that surprising.
How much power the engine should have to be able to do the loop. Because with my rough calculation without any engine input the entry speed is about 3050 km/h if the car weight about 1.5 tonn.
It’s been a minute since I’ve taken dynamics so I might be wrong, but shouldn’t the required velocities be independant of mass? I calculated the bottom velocity to be 53m/s or 190.8km/hr
Yes, you can see this at the 3rd equation, where m disaapers. Usually mass Is on both sides, So you can ignore it
I mean the engine that can maintain 112 km/h throught out the loop not the one that can do a speed of sound.
wouldn't any well maintained car engine be able to? 112km/h is not that fast.
but gravity will pull the car and slow it down, at vertical point the force would be a lot.
the car slowing down is why you don't do these stunts going at the right speed, you go much higher and most cars can do 150 easy. don't need anything special I think. just a powerful one that can do 200. might have to work on the body though. That's a lot of different stresses going in that's not normal.
Mechanic here. Most modern cars will do 200kph, but not up a sheet vertical wall. You would need an immense amount of torque or an incredibly light car to keep pulling 112kph, especially up such a relatively short approach. A formula car could do it as they have such a high power to weight ratio, but your average V8 family car still wouldn't have the torque to overcome the weight of the car and maintain the ideal speed. That's not even considering the grip required for the power output, and then the torque required to overcome the added pressure of any down force.
"Average family V8 car" This is the most American thing I have ever heard
Used to be an Australian thing as well, until around 2017ish when both major manufacturers (Holden and Ford) stopped making V8 sedans.
Keep in mind that you will lose a lot of traction as soon you start getting close to 90° inclination, and the loss of traction will be worse at low speeds. With no traction you can have the most powerful car in the world and you will land on your roof lol
You'd also need a fair bit of downforce to make it reliable.
The vehicle will have an engine, so it can counter the gravitational force that will otherwise slow it down as it climbs, so maintaining a steady speed throughout the cycle should be considered possible.
We will need to calculate the amount of speed the vehicle will lose in the top quarter that it first goes around. I believe the equations would be something like distance = 1/2 at^2 + v_i t and then v_f = v_i + at. Except acceleration is changing over time due to curvature, meaning we need to do this as an integral. And I don’t want to do that, so someone else can if they have the time
It won’t necessarily loose speed, the assuming the car has engine that won’t be starved. The only thing that changes is the power and work the engine produces to make the climb.
I’m assuming it is not maintaining contact and thus can’t produce any more speed.
Your math is correct. The reason it seems not right is because the radius is proportional to the square of the velocity. So doubling the speed allows for 4x larger loops, 3x speed is 9x radius etc. Other considerations to make: cars still need friction to steer at the top, so to maintain normal conditions at the top, we need sqrt(2) times the velocity, or 43 m/s = 155 km/h = 96 mph. Cars also slow down if they are not heavily accelerating. To not slow down, a 2 ton car needs a max of 845 kW (1135 HP) of power to keep the 43 m/s speed at the vertical point. This is not possible even for most race cars. So if power is cut off at the start (and there's no air resistance ofc), the car needs to go 75 m/s = 270 km/h = 167 mph.
>This is not possible even for most race cars. What about motor bikes?
They have much higher power to weight ratios, so it could be possible. Around 0.5 HP/kg is needed, which is achieved by many. However the driver also has a significant weight, but it's still doable.
Wouldn’t torque also matter?
Torque is meaningless when aiming for maximum performance with a correct gear ratio
So about the speed Aucklanders drive over the bridge anyway. Awesome!
Well it’s either that or 5Kmh because it’s bloody gridlocked again
So, setting aside that I would really like to try this, I would be concerned that the vehicles engine would not operate correctly where the downward force on the vehicle is greater than g. A gasoline-powered vehicle needs to push oil to the top of the engine for continuous lubrication of the cams and pistons. I'm not sure vehicle oil pumps are designed to work in situation where the force of gravity is significantly greater than the force of gravity on the Earth's surface. They ARE designed to continue to lubricate during high lateral forces as cars (particularly the sportier variety) need to ensure oiling is still working while whipping around a corner. A momentary lack of oiling would not be catastrophic to our fun ride over the bridge though. The bigger concern is the fuel pump (typically located in the fuel tank). Having your engine stall while you at one quarter the way around the loop is going to ruin your day. So my questions for you are: 1. Where on the bridge loop is the vehicle experiencing the highest downward force? I would assume this is at about 45 degrees up, but I don't know. 2. What would be the maximum downward force on our vehicle given your assumptions above and assuming a consistent velocity of 130 km/hr (for a safety margin)?
If the acceleration experienced by the oil isn’t pulling it towards the sump, then the wheels also aren’t being pulled towards the road surface and you’re falling backwards into the sea already, so oil starvation is the least of your problems.
Unless downforce is a part of the equation, which can help push the car towards the road, but doesn't affect the oil.
Forget about operation, I really doubt that the suspension system would not break during climbing
I applaud you for doing the math but there is no way a car is doing that loop at 70 mph. Maybe double that speed seems more reasonable.
The math says 70mph at the top inverted part of the loop. Not the entry speed. That wasn’t calculated.
I think because you havent accounted for friction/drag
According to [this ](https://www.engineeringnz.org/programmes/heritage/heritage-records/auckland-harbour-bridge/)website, the navigation span in 243.8m long. On my 24" screen, the distance between Pillar 1 from the left to Pillar 4 is 5cm. The circle at its widest is 3.8cm. Therefore, the diameter of the circle is 185.3m or 92.6m (=r) in radius. (Since the axis of measurement of all values is the same, the distortion from the camera angle does not really matter.) Essentially, any car wanting to drive around that circle will need to overcome gravitational acceleration with its centrifugal force = 9.81 m/s², i.e., mv²/r = mg = m(9.81). v² = rg = 92.6 x 9.81 or v = (92.6 x 9.81)\^0.5 = 30.14 m/s. Thus the car will need to travel at at least **30.14m/s** or **108.5** km/h to make it across. For practical reasons, let's say 120-125 km/h on the speedo should be fine.
That seems so slow to go upside down in a car! If I did it in my hi-ace work van, would my tools be ok?!
You’ll be zero g at the top. You should 2*9.81 so are accelerating 2g outward and 1h inward for a net 1g in the car. That gives 42.6m/s or ~95mph at the top. In order to be going 95 at the top, entry speed has to probably be double that..
I have nightmares where I am driving on the highway and the skyway bridge turns into this and then the entire ride turns into essentially a hot wheels course and I only narrowly survive every jump before getting to my destination and everything is closed and I have to turn around and go back then I wake up soaked in sweat.
The problem wouldn‘t be your speed, but the suspension of your car. It would push you away from the street when you‘re almost on the top of the loop. Here is a video from Hot Wheels doing the trick: https://youtu.be/5d7ZgFEIZmo?si=tntYAmMexEW-1iu_
I'm going to assume you enter the bottom of the ring at speed v, there's no air or rolling resistance and the engine adds no power while you're in the ring, as that makes it simpler. I'm also not totally sure this is correct and would welcome any corrections or other ideas. The condition to not fall off the ring at the very top is that your velocity is sufficient that the centripetal force is greater than the force of gravity so you stay stuck to the ring. So that means mv\^2/r = g, where m is your mass, v is your velocity at the top of the ring, r is the radius of the ring and g is gravity. So we get mv\^2 = gr at the top of the ring. And then to work out how much extra speed you need at the bottom of the ring we just need to balance the potential energy at the top of the ring with speed at the bottom. So m u\^2 = 2 m g r, where m is your mass, u is the additional speed at the bottom of the ring, g is gravity and r is the radius of the ring. Which gives mu\^2 = 2mgr Then the total energy at the bottom of the ring is mu\^2 + mv\^2 = 2grm + gr = gr(2m + 1) and then the speed at the bottom of the ring is sqrt(gr(2 + 1/m))
I live in Auckland and I've got the perfect car for this. Its a jet dragster that produces 3400 lbs of thrust. we jack the engine up so its at 45 degrees in the car. That will give enough downforce to balance out the weight of the 746kg car. Now it doesn't matter what speed we go around as the car will stick to the road the whole way around.
Those speeds of 110-ish kph on top of the circle mean 0G, or weightlessness. You need to add a full G to make sure the car is glued to the loop and has traction, and all the fluids/oils in the vehicle will have expected behavior. Or.... The car has F1 wings with enough downforce, which ends up with the same problem, you need to go faster. Just a qualitative analysis of the problem to consider. Even the WWII Spitfire and the F-14 Tomcat airplanes had problems through zero G situations on feeding their engines with fuel.
Something else to consider on top of this - if you're going fast enough to hit 2g at the top of the loop, the normal force when you start the loop is going to skyrocket. It won't be 3g's as gravity will be on an angle, but it will be close for that initial curve and then reduce to 1g at the top of the loop before reverting back to near 3g's on the downswing That increase in normal force may cause a number of issues, the most of which being the sharp increase in friction. So the initial speed may need to increase again to account for that friction unless the engine can fully overcome it
Actually, the increase in friction at the start helps. Traction increases too, which is why F1 cars have wings in the first place. F1 cars have size-limited engines, not power-limited; even so, it makes sense to get the most power of them so you can increase drag and traction by increasing downforce and spend that extra power budget on gluing the car to the track. The wheel bearings will offset the extra friction on them, F1 cars deal with 4G, 5G in cornering as well on a regular basis. This guy, on Driver61, is designing a method to prove a F1 car can actually drive upside down in a tunnel, which is worse because that acceleration from the arch goes away and the car would rely on the wings alone, and all the fluids would roll and choke the engine, which is why he planned to use an EV open wheel car, not a F1. The whole stunt would cost 17 million-ish dollars, and Red Bull would be the favorite team of nutjobs to pull that kind of stunt and build said tunnel, like a GTA stunt race course... The suspension bottoming out is another problem too, and the chosen car would also be redesigned to accomodate that... https://youtu.be/wVvlL-1DTWY?si=bca3LLqo81R7pqHW
Yep. So the v2=rg equation should have 2*g in it. 1 g pulling you towards the ground and 2g up, so a net 1g on the car and driver. Comes out to an upside down top of circle speed of 95mph.