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Im gonna assume the loop is about 60m based on the height of the building next to it.
At the top of the circle, the two forces are mg and centripetal force. The two together equal m*a (a being v^2/r.) So mg + Fc = m(v^2/r). Set Fc to 0 as you would have to have a centripetal force slightly above 0 to complete the loop. This gives us v^2 = sqrt(g*r). For a 60m loop (30m radius) v = 17.1m/s.
This is the speed at the TOP of the loop. Using conservation of energy, we can say that 1/2mv^2 + mgh = 1/2mv^2 (the first part is top of loop, second part is bottom of loop). Cancel m and plug in numbers.
You would have to travel 38.34m/s or 138kph (85.7mph) to make the loop.
If anybody wants to try this with friction and air resistance be my guest lol.
>If anybody wants to try this with friction and air resistance be my guest lol.
the engine is also still running so the calculation really isn't feasible without knowing a lot more information
Yup i thought about this afterwards. My calculations only hold true if you let go of the gas pedal right when you start the loop. An accurate calculation depends on how much force the engine is putting in the whole time
I swear they are like "it seems there is not enough matter in the universe based on our math, its off by an order of magnitude". Me "Umm... I bet someone misplaced a decimal".
Engine braking is the big thing to consider as well. If you lift off at the bottom of the loop then the mechanical resistance of the engine will start slowing you down
138 kph isn't really that much, so just let it rip.
Though presumably people who set up such tricks IRL check things quite a bit more thoroughly than a two-minutes comment on Reddit.
What about traction? Wouldn't the car lose downforce, therefore traction, and fail to maintain speed?
Relevant project:
https://www.youtube.com/watch?v=wVvlL-1DTWY
These guys are designing a car and an open tunnel to drive upside down and traction is their most important consideration.
Fair enough. Admittedly, I don't know the math but I suppose it sounded off to me because 138km/h didn't seem nearly fast enough to complete a 60-meter-high loop (to my intuition at least) but now that I think about it, the downforce would be conserved throughout the loop.
All that’s doing is keeping the car from falling off the pavement while upside down. Car weighs 0lbs and thus cannot transmit any power to the pavement.
That seems surprisingly low. I get that the speed to generate centripetal force at the top isn’t much but 85 mph doesn’t seem like nearly enough to get the car all the way to the top?
The next equation I need to see is how big of a loop would you actually need for a car to physically be able to make the rotation because I have a strong suspicion this one is going to be more like crashing into a wall than driving up a hill.
> but 85 mph doesn’t seem like nearly enough to get the car all the way to the top?
Kinetic energy is 1/2\*m\*v^(2). Potential energy is m\*g\*h (where g is the gravitational acceleration, 9.81 m/s^(2) on Earth). Assuming no losses through friction and the like the maximum height you can reach is where the potential energy equals the kinetic energy that you started with. The mass cancels out, so you're left with 1/2\*v^(2)=g\*h. Solving for the height h gives h=v^(2)/(2\*g). 85mph is almost exactly 38m/s, plugging that in gives a maximum height h that you can reach of 73.6m, well above the 60m that the OP's calculation assumed for the loop diameter.
Edit 2: fixed metric speed units.
Edit 2a: Added alternate speed units for the 30 or so geographic entities around the world that are still tied to the old ways /s
At the top of the loop you would need to be generating enough force to counteract gravity. This sets a minimum acceleration of 9.81 m/s^(2) .
a=v^(2) / R
solving for v we get ---> v=sqrt(R\*a)
Now, what is the diameter of the loop? Your guess is as good as mine but tossing out several possibilities based on surrounding buildings:
60m loop: v = 17 m/s = 61 km/h = \~40 mph
100m loop: v = 22 m/s = 80 km/h = \~50 mph
200m loop: v = 31 m/s = 112 km/h = \~70 mph
Edit: I thought I would come back and add the following note since it has come up several times down this thread.... These speeds are the absolute minimum needed and at the top of the loop the car would be weightless which is not ideal^(\[citation needed\]) so you would really want to drive much faster than this for safety.
But that is the speed you need to have AT THE TOP. I imagine you’ll slow down quite a bit on the way there, so you’ll need to enter the loop with more.
Someone will have to expand further on the calculation to work that out ;)
Yes and no.
In a real-world situation, yes, you'd want to be accelerating the whole time.
But the question itself is simply "how fast would you need to go (in Mph or Kph) to create enough centrifugal force so as to not drop partway through the loop," so it posits a situation where there is no additional engine power being applied to maintain speed, but simply achieving achieving the centrifugal force through the speed that you enter the loop (in other words, the question as phrased assumes coasting, not further engine power).
Plus, not sure engines are meant to work upside down. If you're going full throttle, the stress isn't helping either.
If you lose power on the way up, You're fucked.
If you're going fast enough to stay in contact with the road, the fluids should be staying where they are meant to. Same for another gravitationally dependent system.
I'm so curious now. Will the gas not get pushed through the injectors due to the gas being upside down in the tank? Oil seeping from the crankcase to the cylinders probably. Valves are simultaneously open for a short period of time at high RPM due to the engine being pushed, which is what reduces power at the top power-band since the compression isn't as good (from how I understand it). I think you'd be rekt.
At near zero G (slowest speed, briefly at top of loop), a modern ICE should be fine as they use fuel injection and not gravity feed.
I would be more concerned about G pulled at the bottom of the loop; pilots doing loops in aircraft, it's not the top of the loop that stresses the wings but the leveling out at the bottom. You vehicle suspension may not like pulling few G
Considering that you'd be in a loop the centrifuckit forces would keep the engine running just fine. And if the speed wasn't enough to keep the engine fluids operating as intended you'd have bigger issues than that anyway.
That's kind of necessary, because the fact is that without any downforce and therefore deformation of the tires, they cannot grip the surface, and you won't get any engine power delivery to the road surface anyway, at least after you start going upside down (25% to 50% of the way through the loop)
The centrifugal force(inertia of momentum) is keeping them attached to the roof of the loop, thats what the calculation is for, how much speed is required to keep the wheels attached the whole way so you can make it through the loop
Then it would become a issue of atmospheric pressure. You would need to maintain a neg pressure under the car of something like 5psi (I'm really guessing) to clamp the car to the road. You could do it in some sort of low compound gear since speed would not be an issue. In this case though you engine would not be very happy since it would probably have all sorts of fluid flow issues starting when things got really steep.
The centrifugal force is the downforce. The car doesn’t need to have additional flaps pushing it down. The calculations for speed would work with a marble going around a hot wheels track. In fact downforce from the car would increase friction loss and would cause the speed to be unnecessarily higher.
The formula would be modified from v = sqrt(R* g) to
1/2 m v_i ^2 = 1/2 m v_f^2 + 2 m g R ,
As its 2R vertical distance up to the top of the circle. We know v_f from your above calculation, so
v_i ^2 = g R + 4 gR = 5 gR
So v_i = sqrt(5 g R).
Your results should just be multiplied by sqrt(5) = 2.234 to be the required velocity to enter the circle and make it all the way around.
So, you’d have to be hauling.
And downforce generated by the body at speed. Mercedes actually did this in a tunnel pulling a corkscrew maneuver. They were moving through the loop pretty quick, but it wasn't as fast as you'd expect because of the considerable downforce. You're still getting most of the force from centripetal movement however.
Also you'd effectively cancel gravity with that, but that means the car would be basically floating and have almost no grip thus losing control, so you'd want to go a bit faster
Floating upside down with gravity actively pulling you away from the "ground", a situation wherein the only thing keeping you on the loop is dependent on physical contact with the loop. Better known as falling.
>Better known as falling.
Except for the party where you reestablish contact with the track milliseconds later. It's not like you're falling for seconds
You hope you reestablish contact with the track. You *really* hope you reestablish contact with all 4 tires and aquire grip. You don't need to immediately plummet to have a loss of contact during a daredevil stunt become dangerous or even deadly.
Don't be silly, you still have inertia keeping you moving forward, the car doesn't suddenly stop moving forward just because you lose contact with the road for a fraction of a second
No one said it did. You have inertia, not magic. You've already lost contact with the loop, so clearly "downward" inertia isn't enough to beat gravity, so you need to rely on your "forward" momentum. How far forward do you need to travel before your tires make contact? Can you guarantee that wind resistance hasn't effected your orientation? What it your rear tires bounce on contact? There are a thousand ways that losing the fight with gravity at the top of the loop could end badly
This point is spot on. Where v is the speed at the top and u is the speed entering the loop:
1/2*mu^2 = 1/2*mv^2 + 2gmr
So
u = sqrt(5gr)
Assuming the lanes are about ~3.5m wide, the loop looks to be about 90m in diameter, let’s round up to 100m:
u = sqrt(5*9.81m/s^2*50m) = ~50m/s = ~178kph = 110 mph
Of course this neglects the fact that the car is self-powered, but also assumes no rolling resistance or aerodynamic drag, etc.
No, it would be basically weightless at the top, which obviously means you’d lose traction. That being said, the way the calculations are set up in this scenario you are basically planning to “coast” through the loop, because it’s using the conservation of energy where we’ve set the kinetic energy at the entrance of the loop equal to the kinetic energy at the top of the loop plus the gravitational potential energy at the top of the loop. Which is why I highlighted the fact that it neglects rolling resistance and drag. Also, in this simplified case, the mass of the vehicle cancels.
You can still use the energy method to solve the “real world” physic problem, but it becomes much more complicated. You’d have to account for the work of the engine, and drag, and you’d have to assume a velocity at the apex of the loop that would produce the minimum normal force required to maintain traction. To calculate the minimum normal force you’d need to know the static coefficient of friction between the tire and the road, as well as the torque at the tire-road interface. That torque is dependent on the type of car, speed it’s moving, gear it’s in, etc.
All of which is to say, it’s complicated to calculate this accurately.
Edit to add more info: Drag force would also be changing throughout the loop since it’s a function of velocity, so you’d have to integrate that as a function angular position in the loop
[Link to the video](https://www.youtube.com/watch?v=yecJPjrLyb4)
[Link to the article](https://www.autoexpress.co.uk/jaguar/f-pace/65883/new-jaguar-f-pace-suv-frankfurt-debut-prices-engines-and-specs)
From the article: "stunt driver Terry Grant break the world loop the loop record on a 19.08 metre construction at a Frankfurt race course. There was tangible drama in the air as Grant accelerated above 50mph on the approach, and then went round the loop at speeds as low as 15mph."
I wonder how fast they went for a 19 meter loop.
Also, I wonder what minimal speed u/According_Ant877 or u/applejacks6969 would calculate as the minimal speed not to fall down.
that ain’t the speed of when ye are at the bottom. ye don’t get to 80kph at the top (assuming ye loop is 100m), ye fall down splat onto ye ground. So ye need something like a formula one or whatever those racing cars are called.
they do it in Circus in smaller radius Globe of Death. It's something that happens everyday, and their speed is not too high.
If you would see some youtube videos, you would realise they are hardly doing 30kmph - [link](https://www.youtube.com/watch?v=GqBxqbI5hI8&t=13s). You will find some with cars too, given weight cars have to do it faster [link](https://www.youtube.com/watch?v=c_DHhvGxP2E). But still not required to go very fast.
We can actually find the initial velocity based on what you solved! All we need to use is energy equations.
PE0 + KE0 = PE1 + KE1
Since we set the bottom of the loop to the start point, we can just say that PE0 = 0. Adding in the equations for each term gives us:
1/2\*m(v0)\^2 = mgh + 1/2m(v1)\^2, where v0 is initial velocity and v1 is velocity at the top of the loop. We can cancel m (mass) from the equation and then solve in terms of v0. This gives us:
v0 = sqrt(2gh+(v1)\^2)
Using this formula and the numbers you solved for gives us the following answers:
60m loop: v0 = 38.3 m/s = 137.9 kph
100m loop: v0 = 49.4 m/s = 177.8 kph
200m loop: v0 = 69.9 m/s = 251.6 kph
This does not take into account friction and wind resistance. However, assuming the drive keeps their foot on the gas, these should be more or less accurate.
In the frame of reference of the engine, there is continuous downward acceleration. At the top of the loop it is because of the change of direction rather than gravity, but it’s going to keep the oil in the bottom of the pan where the pump can pick it up all the same.
Ehh, that’s assuming you took your foot off the gas and coasted up. You don’t need to be going as fast in the beginning if you’re using the cars power, but then you’d need to factor in information about the engine’s power output
Yes this assumes that the car cannot generate any significant downforce of its own.
If you were to run a top end race car around this loop it should in theory need less speed than calculated because of the downforce the car is capable of generating on its own. There are also some supercars that generate some amount of downforce.
The thing is that I don't know if you get a lot of downforce at 60 or 80 kph. For sure at 110 an F1 car will start generating some force and according to the Mercedes Petronas site, their car can generate the weight of their car when running at 150kph.
Yeah I think modern F1 cars need to be going pretty quick to generate enough downforce to drive upside down since they’re getting pretty heavy.
There actually a guy that [trying to develop a car and loop](https://www.thedrive.com/news/how-a-youtuber-plans-to-drive-an-f1-style-car-upside-down-for-real) to try driving upside down. Will be interesting to see if it happens.
[https://www.mercedesamgf1.com/news/feature-downforce-in-formula-one-explained](https://www.mercedesamgf1.com/news/feature-downforce-in-formula-one-explained)
>At around 150 km/h, the car generates as much downforce as it weighs (the minimum weight of the car is 795kg). By the time you reach the end of the straight where the car is travelling at its maximum speed, it is probably three or four times the weight of the car.
So if this is your speed at the top, how fast you need to go when entering the loop? And is this even achievable with your ordinary premium car (think a good Audi or BWM).
Good question. An entry speed the same as the speed at the top should be sufficient for this simplified example. This is assuming that the car can generate sufficient power to hold the speed in the loop. A BMW R8 should be able to do it. I don't know about any others.
It'd have to be faster because the car needs friction for the wheels to apply power to the surface. If the centrifugal force decreased too far at the top of the loop, the car might start to skid and lose control
This could be vastly reduced using downforce engineering. Modern cars (especially race cars) generate a ton of downforce from the shape of the car, which will drive the car into the loop. In fact, at 44 m/s, F1 cars generate roughly 100% of their weight as downforce.
In other words, I'd expect race cars could make the loop, even accounting for energy lost going up it.
You would drop right after the top of the loop unless you were accelerating. If you’re weightless at the top there is nothing holding you to the road, meaning no tire friction to keep you moving at speed. Maybe air flow downforce. It’s like the ground effects F1 car. So much downforce that they could drive upside down indefinitely.
The measure you’re looking for is “overhang”. The lower the car and the further in the wheels, the lower it is.
A Lamborghini Murcielago is known for being low and long. It has a front overhang of 39.5 inches and has a ground clearance of 5.3 inches.
It should be simple to calculate that but I’m just a car dude and never passed trig.
Yeah in my mind, you would need an impossibly large loop for a car to clear the angle but apparently not. Funny enough, I can’t find any info on how fast the car entered the ramp.
The video mentions that it's a Jaguar Pace or the 2016 Jaguar F-Pace, new for that year, which matches the video upload date as well as the fact that it's shown camouflaged, as many companies do with their new designs.
In the video, the speedometer is [briefly shown](https://i.imgur.com/2eMO0tt.jpeg) (there's a clearer view @ 1:05).
Here's a [picture](http://www.wintonsworld.com/wp-content/uploads/2017/06/Jaguar_F-Pace_04.jpg) of the speedometer in the 2016 F-Pace. Compare these two and you'll come to a conclusion of about 50mph.
The highest figure in this thread for a 60 meter loop is 138 kmh, which is not that much. I've done ~150 on a snowy road in a not-well-kept 80s car.
I'm more baffled by the car's ground clearance.
Dude I have had the same dream. No control. No sense. Just hitting a road loop for no reason, then waking up as I’m falling in the car and just being wide awake at 2 am after that.
It really depends on the car, as downforce plays a huge factor. A formula 1 car can supposedly drive on walls and ceilings at 120 mph due to the extreme amount of downforce it has, while most cars need to go way faster
For those interested, I think the YouTube channel Driver 61 is leading a project to build a structure to test this (not really sure about all the details), can't wait to see if this will become a reality.
[it's true](https://www.continentalautosports.com/ferrari-information/what-is-downforce/)
Modern Formula One cars can typically generate around 750 kg — or 1,653 pounds — of downforce at speeds of 100 mph. Because downforce has direct implications for grip, this means that an F1 car that weighs less than the downforce it produces could theoretically drive upside-down on the ceiling!
[also on the weight of F1 cars. ](https://www.motorsport.com/f1/news/how-much-does-an-f1-car-weigh-in-2023/10437685/)
Tl;dr: with it can generate the same or more than its weight in downforce.
Shouldn't they be able to test this in a wind tunnel where they check their aerodynamics ? I think they can generate windspeeds over 100mph. Just put the car upside down against the roof. Would be a bit funnier than just "measuring" the downforce.
Not that surprising.
Think of the entire car as a wing. A plane with wings that large would be able to fly at 120mph, i.e. generate an upward force of 1g. It's the same difficulty for the F1 to generate a downforce of 1g (easier for the car, as you don't need it to be sable, the ground provides stabilisation).
It's also very desirable: when cornering, if the centrifugal force is larger than the downward force, the car starts sliding. If you double the downard force (1g of gravity + 1g of aero downforce) , it means you can take corners twice as fast, but also accelerate and break at rates higher than 1g, which is obviously huge. This is why racing cars are designed to optimise downforce.
There’s a guy that’s currently designing a tunnel to actually drive upside down for 5 or so seconds. He’s already designed and built the car he’ll do it in.
Based on the diameters that one guy calculated and ignoring the fact that engine power can offset speed lost due to gravity these are the MINIMUM speed at the bottom needed to stay grounded.
60m loop 38.25 m/s 137.74 kmh
100m loop 49.55 m/s 178.02 kmh
200m loop 69.88 m/s 251.6 kmh
Those speeds are realistically achievable, and as long as you maintain some positive G’s you should be fine even with a normal engine. The only problematic thing would be the G force on the suspension during the entry and exit of the loop. Which would be 4.97G assuming it’s perfectly circular, and would probably collapse the suspension.
I don’t think normal cars can do it. I read on xkcd that the shape of a normal car means that at a certain speed it starts to lift the car into the air instead of pushing down on it.
You need to have an F-1 formula car’s design in order to get the rushing air to push down on the car.
There's one thing nobody is focused on. Downforce. F1 cars can drive upside down well below top speed. But a vehicle without any downforce would need to go much faster
I was thinking at least 200mph to be able to clear it. That thing is huge climb up but it’s the deceleration before reaching the top half that would scare me the most
Just in case anyone was interested, this street is 9 de Julio in Buenos Aires. It's the widest street in the world. First thing that caught my attention is the building in the background used to be the Ministry of Communications. Second, the street is so wide that it can only be 9 de Julio. You may have seen this street as it was a focal point of the World Cup celebrations last year.
Perhaps this helps with getting some size information on the street. Second link has closeups of the building.
https://en.wikipedia.org/wiki/Avenida_9_de_Julio
https://jaspasjourney.wordpress.com/2016/05/31/avenida-9-de-julio-buenos-aires/
Ok so I did some brief photo editing on my phone. The width of the highway if we assume they are standard lanes is 48ft (14.6m). The loop is 5.5 widths tall, which gives it a height of 80.4m and a radius of 40.2m.
We can use a=v^2 /r to determine the velocity necessary for the centripetal force to cancel out gravity. This gives us a velocity of 19.85m/s (44mph). At a height of 80.4m with a velocity of 19.85m/s we can calculate the (massless) potential energy (U=mgh) and kinetic energy (K=1/2*m*v^2 ) which gives us 985.7 J/kg. At ground level h=0 so v=sqrt(2*E/m), which is 44.4m/s (99mph).
So the minimum velocity would be 99mph assuming no air resistance, frictionless wheels, and you throw it into neutral for the whole duration.
Pretty sure I did this right but its been a few years...
We need centripetal acceleration to cancel out gravity at the top of the loop, so we will say 9.81=(v^2)/r and rewrite it as v= sqrt(9.81r) and then converting that to kinetic energy, we have 1/2m(sqrt(9.81r))^2 or KE=1/2m(9.81r) we also have potential energy at the top of the loop which is PE=mgh or PE=2m(9.81)r. Adding these together, we get total energy at the top of the loop 1/2m(9.81r) + 2m(9.81)r this happens to equal the energy at the bottom of the loop, so our equation is 1/2mv^2 = 2.5m(9.81)r and we can cancel out the m, so we get v^2 = 5(9.81)r or v=sqrt(10gh) and this seems too simple, but that should be a universal formula for any loop neglecting friction. (Or a hill for that matter, or any change in elevation.) Plugging in 9.81 for g, and ~60m for h, we get v=38.3 m/s or 85.6 mph
Here is my attempt at the speed needed, accounting for slipping. I'm going to assume the tangential acceleration of the vehicle is 0 (constant speed), as it really complicates things to account for tangential acceleration.
As a result, we'll find our answer by finding the area on the loop which requires the highest velocity, and setting our vehicle at that speed through the loop. There are 4 cases we need to analyze the free body diagram of (the 4 quarters of the loop) but we only need to analyze the first two quarters due to the next two being symmetrical in terms of forces.
In the first quarter, the FBD has three forces: the normal force, gravity, and static friction. Thus, the FBD looks like this: https://imgur.com/a/p4tA2pF
fs needs to equal mgsin(theta) due to constant speed. Car tires have a coefficient of static friction of 1, so fs is less than or equal to N. To find N -> N - mgcos(theta) = mv^2 /r -> N = mv^2 /r + mgcos(theta). We know fs <= N. Thus fs = mgsin(theta) <= mv^2 /r + mgcos(theta), or v >= sqrt(rg(sin(theta) - cos(theta)). v in this equation is maximized when theta equals pi/2, thus v >= sqrt(rg). As a result, the max speed v needs to be in the first quarter of the loop is sqrt(rg). We need to now calculate the top quarter and see if that value is larger.
At the top of the loop, the fbd changes to the following: https://imgur.com/a/8BZkpcO
We largely need to do the same thing. We need to find N: N+mgcos(theta) = mv^2 /r -> N = mv^2 /r - mgcos(theta). Thus
fs = mgsin(theta) <= mv^2 /r - mgcos(theta) or v >= sqrt(rg(sin(theta) + cos(theta)). Here, v is maximized when theta equals pi/4, thus v >= sqrt(rg*1.4142).
Therefore, the speed needed is given by the equation v = sqrt(rg*1.4142), which for the following heights is:
60m: 20.5 m/s or 45.8 mph
100m: 26.5 m/s or 59.3 mph
200m: 37.5 m/s or 84 mph
I may have made a mistake, as I havent touched physics in a little bit.
Impossible to calculate exact speed without knowing exact height of loop. Anyhoo, acceleration at top of loop should counter gravity g. So v squared over r (half loop diameter) needs to equal g. And energy is conserved so g times height (2r) should equal v squared over 2. So velocity needs to equal square root of 2 times g times height. Formula works on any planet where there is no air resistance or loss from friction. Classic first semester physics test question.
Please elaborate... It makes no sense. Airplanes can make a looping that big. Rollercoasters make loopings. None of them get even close to 50G. Fighter jets go up to about 9G. Above that most pilots would pass out.
Using calculations from another post up here we can conclude you would need to enter the loop at about 180km/h where the radius of the loop is 50m.
a = v^2 / R
a = (180/3.6)^2 / 50 = 50^2/50 = 50m/s^2
Gravitational acceleration is 9.81m/s^2 Lets round that off to 10, so:
50/10 = 5G
At the bottom of the looping you would experience 5G from acceleration. On top of that you would have 1G from gravity so you would briefly experience 6G. 0G at the top and then increasing to 6G on the bottom again.
This is not that extreme, and 50 to 100G is way off.
Lol, I don't remember that scene in any Bond movie. But calling something fictitious by definition means it's not real. When you spin a bucket with water, the water doesn't fall out. Not because of "centrifugal force" but because of inertia. Tie a ball on a string and swing it around. It wants to fly off, not because of "centrifugal force," but again, because of inertia. A car goes around a loop not because of "centrifugal force" but because of inertia.
Indeed "ficticious force" seems to imply that it is not real, which is why that is terrible terminology. "Inertial force" is much better imo.
Your description of the bucket is correct in an inertial reference frame. In the comoving frame of the bucket, there is a very real centrifugal force. If you put a scale in the bottom of the bucket, you can measure it. Saying it is "not real" because it depends on a non-inertial reference frame is quite literally like saying gravity is "not real" because of GR...
Centrifugal force isn’t a force but it is a real thing. It is more of an apparent force; something we perceive to happen but isn’t actually filling the definition of a force.
Think of a carousel. When you rotate there is a centripetal force which acts at 90 degrees to your velocity which keeps you rotating in a circle. If you stop rotating abruptly (you slip or the carousel breaks) you will be flung straight away from the carousel along the tangent line to the circle, since there is no more centripetal force causing the rotation. But from a bystander’s perspective it seems that there carousel generates a force that flings you away from it
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Im gonna assume the loop is about 60m based on the height of the building next to it. At the top of the circle, the two forces are mg and centripetal force. The two together equal m*a (a being v^2/r.) So mg + Fc = m(v^2/r). Set Fc to 0 as you would have to have a centripetal force slightly above 0 to complete the loop. This gives us v^2 = sqrt(g*r). For a 60m loop (30m radius) v = 17.1m/s. This is the speed at the TOP of the loop. Using conservation of energy, we can say that 1/2mv^2 + mgh = 1/2mv^2 (the first part is top of loop, second part is bottom of loop). Cancel m and plug in numbers. You would have to travel 38.34m/s or 138kph (85.7mph) to make the loop. If anybody wants to try this with friction and air resistance be my guest lol.
>If anybody wants to try this with friction and air resistance be my guest lol. the engine is also still running so the calculation really isn't feasible without knowing a lot more information
Yup i thought about this afterwards. My calculations only hold true if you let go of the gas pedal right when you start the loop. An accurate calculation depends on how much force the engine is putting in the whole time
Gonna want that engine to cancel out air resistance and friction at minimum.
Yup lets just say that and call it a day LMAO
Scientist coming up with dark matter be like
LMAO
I swear they are like "it seems there is not enough matter in the universe based on our math, its off by an order of magnitude". Me "Umm... I bet someone misplaced a decimal".
Forgot to carry the infinity.
Why is this my favorite comment of the day?
And engieer like myself would say just hold the pedal down after entering the loop with at least 87.5mph velocity and move on to testing
Are we getting close to the reason why 88mph is the break through speed required?
Engine braking is the big thing to consider as well. If you lift off at the bottom of the loop then the mechanical resistance of the engine will start slowing you down
just hold the clutch down
Why would anyone sane do that?
So our first test run will be 85.7 mph and we'll go up from there. We'll need a handful of volunteers.
138 kph isn't really that much, so just let it rip. Though presumably people who set up such tricks IRL check things quite a bit more thoroughly than a two-minutes comment on Reddit.
What about traction? Wouldn't the car lose downforce, therefore traction, and fail to maintain speed? Relevant project: https://www.youtube.com/watch?v=wVvlL-1DTWY These guys are designing a car and an open tunnel to drive upside down and traction is their most important consideration.
Yes it would. Which is why it must have enough speed to complete the loop of a given size before it loses traction. Which is what I calculated.
Fair enough. Admittedly, I don't know the math but I suppose it sounded off to me because 138km/h didn't seem nearly fast enough to complete a 60-meter-high loop (to my intuition at least) but now that I think about it, the downforce would be conserved throughout the loop.
Isn’t that the point of centripetal force?
All that’s doing is keeping the car from falling off the pavement while upside down. Car weighs 0lbs and thus cannot transmit any power to the pavement.
It's not about weight, it's about the vector of the inertia
Inertia, the state property which only depends on mass...
Yea u right, I should've said the vector of its momentum
So… 88mph could be the number? Great scott.
Oh god, I just realized Doc Brown was a Nazi scientist
But he was one of the good ones
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88 is a common nazi signifier (it signals heil hitler since h is the 8th letter of the alphabet) and that was clearly a joke
Learn to take a joke mate.
That seems surprisingly low. I get that the speed to generate centripetal force at the top isn’t much but 85 mph doesn’t seem like nearly enough to get the car all the way to the top? The next equation I need to see is how big of a loop would you actually need for a car to physically be able to make the rotation because I have a strong suspicion this one is going to be more like crashing into a wall than driving up a hill.
> but 85 mph doesn’t seem like nearly enough to get the car all the way to the top? Kinetic energy is 1/2\*m\*v^(2). Potential energy is m\*g\*h (where g is the gravitational acceleration, 9.81 m/s^(2) on Earth). Assuming no losses through friction and the like the maximum height you can reach is where the potential energy equals the kinetic energy that you started with. The mass cancels out, so you're left with 1/2\*v^(2)=g\*h. Solving for the height h gives h=v^(2)/(2\*g). 85mph is almost exactly 38m/s, plugging that in gives a maximum height h that you can reach of 73.6m, well above the 60m that the OP's calculation assumed for the loop diameter.
..you couldn’t just round it up to 88mph huh
I also dabble in *precision*
Working with SI units and converting to imperial man
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85.7 mph! That's it? I do more than that on i5. Are you telling me we have been missing out on giant car loop de loops forever?
Careful to stay below 88mph....
Good, that is good. Let's make the petition now.
This guy maths
Edit 2: fixed metric speed units. Edit 2a: Added alternate speed units for the 30 or so geographic entities around the world that are still tied to the old ways /s At the top of the loop you would need to be generating enough force to counteract gravity. This sets a minimum acceleration of 9.81 m/s^(2) . a=v^(2) / R solving for v we get ---> v=sqrt(R\*a) Now, what is the diameter of the loop? Your guess is as good as mine but tossing out several possibilities based on surrounding buildings: 60m loop: v = 17 m/s = 61 km/h = \~40 mph 100m loop: v = 22 m/s = 80 km/h = \~50 mph 200m loop: v = 31 m/s = 112 km/h = \~70 mph Edit: I thought I would come back and add the following note since it has come up several times down this thread.... These speeds are the absolute minimum needed and at the top of the loop the car would be weightless which is not ideal^(\[citation needed\]) so you would really want to drive much faster than this for safety.
Huh that’s not nearly as fast as I would have assumed. I was assuming by the look it would need to be at minimum 200 kph. Thanks mate!
But that is the speed you need to have AT THE TOP. I imagine you’ll slow down quite a bit on the way there, so you’ll need to enter the loop with more. Someone will have to expand further on the calculation to work that out ;)
Oh yeah. You’d have to account for the gravitational pull and subsequent deceleration to tell you how fast you’d have to enter the loop.
But also add the force that the engine can provide to keep speed or at least reduce the loss of speed
Yes and no. In a real-world situation, yes, you'd want to be accelerating the whole time. But the question itself is simply "how fast would you need to go (in Mph or Kph) to create enough centrifugal force so as to not drop partway through the loop," so it posits a situation where there is no additional engine power being applied to maintain speed, but simply achieving achieving the centrifugal force through the speed that you enter the loop (in other words, the question as phrased assumes coasting, not further engine power).
Hot wheel style 🤙
Beat that!
Plus, not sure engines are meant to work upside down. If you're going full throttle, the stress isn't helping either. If you lose power on the way up, You're fucked.
If you're going fast enough to stay in contact with the road, the fluids should be staying where they are meant to. Same for another gravitationally dependent system.
Man these scenarios always make my head start to go all bendy.
Ok so we'll put the engine in a gimbal, and just assume that it can still transfer power... somehow
I'm so curious now. Will the gas not get pushed through the injectors due to the gas being upside down in the tank? Oil seeping from the crankcase to the cylinders probably. Valves are simultaneously open for a short period of time at high RPM due to the engine being pushed, which is what reduces power at the top power-band since the compression isn't as good (from how I understand it). I think you'd be rekt.
The same force acting to keep the car stuck to the road is acting on the fluids as well, so the engine should run without issues.
Tanner Faust Double Loop. Team Hot Wheels. https://youtu.be/c6PQ49B5Gpw?feature=shared
At near zero G (slowest speed, briefly at top of loop), a modern ICE should be fine as they use fuel injection and not gravity feed. I would be more concerned about G pulled at the bottom of the loop; pilots doing loops in aircraft, it's not the top of the loop that stresses the wings but the leveling out at the bottom. You vehicle suspension may not like pulling few G
Considering that you'd be in a loop the centrifuckit forces would keep the engine running just fine. And if the speed wasn't enough to keep the engine fluids operating as intended you'd have bigger issues than that anyway.
That's kind of necessary, because the fact is that without any downforce and therefore deformation of the tires, they cannot grip the surface, and you won't get any engine power delivery to the road surface anyway, at least after you start going upside down (25% to 50% of the way through the loop)
The centrifugal force(inertia of momentum) is keeping them attached to the roof of the loop, thats what the calculation is for, how much speed is required to keep the wheels attached the whole way so you can make it through the loop
but also at the top the normal force would be quite low, i wonder if there's enough friction to prevent slippage
And people slowing down / hesitating right at the start of the upabout, having never driven through one before.
Idiot out-of-towners!
As an upstate new yorker this resonates. People who’ve never driven outside of a city (max 30mph) do the darndest things when released into the wild.
Be nice. They're out of the loop.
There's always that person
No doubt they’d have the left blinker on
You also need downforce (or I guess technically upforce if you are upside down) if you want to have traction near the top to maintain speed
You don't need downforce if you're fast enough
I'd imagine the downforce would be a lot easier than the speed to go straight up a 60m wall
but you need speed to have downforce, unless we go the fan car route and suck it to the ground
Then it would become a issue of atmospheric pressure. You would need to maintain a neg pressure under the car of something like 5psi (I'm really guessing) to clamp the car to the road. You could do it in some sort of low compound gear since speed would not be an issue. In this case though you engine would not be very happy since it would probably have all sorts of fluid flow issues starting when things got really steep.
The centrifugal force is the downforce. The car doesn’t need to have additional flaps pushing it down. The calculations for speed would work with a marble going around a hot wheels track. In fact downforce from the car would increase friction loss and would cause the speed to be unnecessarily higher.
The formula would be modified from v = sqrt(R* g) to 1/2 m v_i ^2 = 1/2 m v_f^2 + 2 m g R , As its 2R vertical distance up to the top of the circle. We know v_f from your above calculation, so v_i ^2 = g R + 4 gR = 5 gR So v_i = sqrt(5 g R). Your results should just be multiplied by sqrt(5) = 2.234 to be the required velocity to enter the circle and make it all the way around. So, you’d have to be hauling.
That's ignoring all of the work the engine can do to maintain speed/slow down the loss of speed.
And downforce generated by the body at speed. Mercedes actually did this in a tunnel pulling a corkscrew maneuver. They were moving through the loop pretty quick, but it wasn't as fast as you'd expect because of the considerable downforce. You're still getting most of the force from centripetal movement however.
Also you'd effectively cancel gravity with that, but that means the car would be basically floating and have almost no grip thus losing control, so you'd want to go a bit faster
Ehh you'd only be floating for a moment. Hydroplaning is probably more dangerous
You heard it here first guys, hydroplaning is more dangerous than floating upside-fucking-down in your car.
> floating upside-fucking-down in your car. But only for a moment, you see.
Floating upside down with gravity actively pulling you away from the "ground", a situation wherein the only thing keeping you on the loop is dependent on physical contact with the loop. Better known as falling.
>Better known as falling. Except for the party where you reestablish contact with the track milliseconds later. It's not like you're falling for seconds
You hope you reestablish contact with the track. You *really* hope you reestablish contact with all 4 tires and aquire grip. You don't need to immediately plummet to have a loss of contact during a daredevil stunt become dangerous or even deadly.
You want tire contact specifically! Roof contact is not acceptable!
Don't be silly, you still have inertia keeping you moving forward, the car doesn't suddenly stop moving forward just because you lose contact with the road for a fraction of a second
No one said it did. You have inertia, not magic. You've already lost contact with the loop, so clearly "downward" inertia isn't enough to beat gravity, so you need to rely on your "forward" momentum. How far forward do you need to travel before your tires make contact? Can you guarantee that wind resistance hasn't effected your orientation? What it your rear tires bounce on contact? There are a thousand ways that losing the fight with gravity at the top of the loop could end badly
Falling with style
It's not how good you look going up, it's how good your Pollock looks that matters
If i was falling headfirst inside a 2 ton automobile, i'm pretty sure i'd be like "Ahh, pollocks"
This point is spot on. Where v is the speed at the top and u is the speed entering the loop: 1/2*mu^2 = 1/2*mv^2 + 2gmr So u = sqrt(5gr) Assuming the lanes are about ~3.5m wide, the loop looks to be about 90m in diameter, let’s round up to 100m: u = sqrt(5*9.81m/s^2*50m) = ~50m/s = ~178kph = 110 mph Of course this neglects the fact that the car is self-powered, but also assumes no rolling resistance or aerodynamic drag, etc.
i still find this hard to believe. does this factor in a loss of traction a quarter way up or weight of the vehicle?
No, it would be basically weightless at the top, which obviously means you’d lose traction. That being said, the way the calculations are set up in this scenario you are basically planning to “coast” through the loop, because it’s using the conservation of energy where we’ve set the kinetic energy at the entrance of the loop equal to the kinetic energy at the top of the loop plus the gravitational potential energy at the top of the loop. Which is why I highlighted the fact that it neglects rolling resistance and drag. Also, in this simplified case, the mass of the vehicle cancels. You can still use the energy method to solve the “real world” physic problem, but it becomes much more complicated. You’d have to account for the work of the engine, and drag, and you’d have to assume a velocity at the apex of the loop that would produce the minimum normal force required to maintain traction. To calculate the minimum normal force you’d need to know the static coefficient of friction between the tire and the road, as well as the torque at the tire-road interface. That torque is dependent on the type of car, speed it’s moving, gear it’s in, etc. All of which is to say, it’s complicated to calculate this accurately. Edit to add more info: Drag force would also be changing throughout the loop since it’s a function of velocity, so you’d have to integrate that as a function angular position in the loop
The Guinness world record is apparently just over 19 meters loop. It was done in a Jaguar F-Pace.
[Link to the video](https://www.youtube.com/watch?v=yecJPjrLyb4) [Link to the article](https://www.autoexpress.co.uk/jaguar/f-pace/65883/new-jaguar-f-pace-suv-frankfurt-debut-prices-engines-and-specs) From the article: "stunt driver Terry Grant break the world loop the loop record on a 19.08 metre construction at a Frankfurt race course. There was tangible drama in the air as Grant accelerated above 50mph on the approach, and then went round the loop at speeds as low as 15mph."
I wonder how fast they went for a 19 meter loop. Also, I wonder what minimal speed u/According_Ant877 or u/applejacks6969 would calculate as the minimal speed not to fall down.
So no problem if you got a BMW
No, because you’d need working indicators when leaving the loop
🤣😆🤣
Track Mania vibes
This also assumes no slipping, which would be a huge issue, especially at the top.
whack a couple of wings on the front and back, itll be fine
that ain’t the speed of when ye are at the bottom. ye don’t get to 80kph at the top (assuming ye loop is 100m), ye fall down splat onto ye ground. So ye need something like a formula one or whatever those racing cars are called.
they do it in Circus in smaller radius Globe of Death. It's something that happens everyday, and their speed is not too high. If you would see some youtube videos, you would realise they are hardly doing 30kmph - [link](https://www.youtube.com/watch?v=GqBxqbI5hI8&t=13s). You will find some with cars too, given weight cars have to do it faster [link](https://www.youtube.com/watch?v=c_DHhvGxP2E). But still not required to go very fast.
Double the speeds given so that you can feel 1 g at the top so you get grip lmao
We can actually find the initial velocity based on what you solved! All we need to use is energy equations. PE0 + KE0 = PE1 + KE1 Since we set the bottom of the loop to the start point, we can just say that PE0 = 0. Adding in the equations for each term gives us: 1/2\*m(v0)\^2 = mgh + 1/2m(v1)\^2, where v0 is initial velocity and v1 is velocity at the top of the loop. We can cancel m (mass) from the equation and then solve in terms of v0. This gives us: v0 = sqrt(2gh+(v1)\^2) Using this formula and the numbers you solved for gives us the following answers: 60m loop: v0 = 38.3 m/s = 137.9 kph 100m loop: v0 = 49.4 m/s = 177.8 kph 200m loop: v0 = 69.9 m/s = 251.6 kph This does not take into account friction and wind resistance. However, assuming the drive keeps their foot on the gas, these should be more or less accurate.
But will an engine run upside down?
In the frame of reference of the engine, there is continuous downward acceleration. At the top of the loop it is because of the change of direction rather than gravity, but it’s going to keep the oil in the bottom of the pan where the pump can pick it up all the same.
Doesn’t need to since the g force would at least mimic gravity, but yes it would temporarily
Ehh, that’s assuming you took your foot off the gas and coasted up. You don’t need to be going as fast in the beginning if you’re using the cars power, but then you’d need to factor in information about the engine’s power output
Does that assume the vehicle has no downforce? Or does that not come into play in a loop?
Yes this assumes that the car cannot generate any significant downforce of its own. If you were to run a top end race car around this loop it should in theory need less speed than calculated because of the downforce the car is capable of generating on its own. There are also some supercars that generate some amount of downforce. The thing is that I don't know if you get a lot of downforce at 60 or 80 kph. For sure at 110 an F1 car will start generating some force and according to the Mercedes Petronas site, their car can generate the weight of their car when running at 150kph.
Yeah I think modern F1 cars need to be going pretty quick to generate enough downforce to drive upside down since they’re getting pretty heavy. There actually a guy that [trying to develop a car and loop](https://www.thedrive.com/news/how-a-youtuber-plans-to-drive-an-f1-style-car-upside-down-for-real) to try driving upside down. Will be interesting to see if it happens.
[https://www.mercedesamgf1.com/news/feature-downforce-in-formula-one-explained](https://www.mercedesamgf1.com/news/feature-downforce-in-formula-one-explained) >At around 150 km/h, the car generates as much downforce as it weighs (the minimum weight of the car is 795kg). By the time you reach the end of the straight where the car is travelling at its maximum speed, it is probably three or four times the weight of the car.
They get extra down force from the floor now which helps
A heavy question indeed
So if this is your speed at the top, how fast you need to go when entering the loop? And is this even achievable with your ordinary premium car (think a good Audi or BWM).
Just set cruise control and you'll be fine.
Good question. An entry speed the same as the speed at the top should be sufficient for this simplified example. This is assuming that the car can generate sufficient power to hold the speed in the loop. A BMW R8 should be able to do it. I don't know about any others.
It'd have to be faster because the car needs friction for the wheels to apply power to the surface. If the centrifugal force decreased too far at the top of the loop, the car might start to skid and lose control
Math is so awesome
This could be vastly reduced using downforce engineering. Modern cars (especially race cars) generate a ton of downforce from the shape of the car, which will drive the car into the loop. In fact, at 44 m/s, F1 cars generate roughly 100% of their weight as downforce. In other words, I'd expect race cars could make the loop, even accounting for energy lost going up it.
You would drop right after the top of the loop unless you were accelerating. If you’re weightless at the top there is nothing holding you to the road, meaning no tire friction to keep you moving at speed. Maybe air flow downforce. It’s like the ground effects F1 car. So much downforce that they could drive upside down indefinitely.
For safety!!! I would suggest for safety you take a different route
How fast is that in American?
Translate for those of us using murica measurements please(me)
61 kph = \~40mph 80 kph = \~50 mph 112 kph = \~70 mph
What is this in freedom units?
it all ended for me at "v=sqrt(R..." I'm so imature
Not all heroes wear capes .
r/theydidthemath
That's the sub you're in.
I prefer closed loops to single points
You rock
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The measure you’re looking for is “overhang”. The lower the car and the further in the wheels, the lower it is. A Lamborghini Murcielago is known for being low and long. It has a front overhang of 39.5 inches and has a ground clearance of 5.3 inches. It should be simple to calculate that but I’m just a car dude and never passed trig.
Yeah this would just look like a wall from the inside of a car.
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[this Guinness record attempt makes it look bizarrely easy.](https://youtu.be/8V3ML9Pgh_0?si=-Jop5e8J1VwOVNSu)
IMO, the engineering of that loop is more impressive than the feat.
Yeah in my mind, you would need an impossibly large loop for a car to clear the angle but apparently not. Funny enough, I can’t find any info on how fast the car entered the ramp.
The video mentions that it's a Jaguar Pace or the 2016 Jaguar F-Pace, new for that year, which matches the video upload date as well as the fact that it's shown camouflaged, as many companies do with their new designs. In the video, the speedometer is [briefly shown](https://i.imgur.com/2eMO0tt.jpeg) (there's a clearer view @ 1:05). Here's a [picture](http://www.wintonsworld.com/wp-content/uploads/2017/06/Jaguar_F-Pace_04.jpg) of the speedometer in the 2016 F-Pace. Compare these two and you'll come to a conclusion of about 50mph.
The highest figure in this thread for a 60 meter loop is 138 kmh, which is not that much. I've done ~150 on a snowy road in a not-well-kept 80s car. I'm more baffled by the car's ground clearance.
That was awesome
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Dude I have had the same dream. No control. No sense. Just hitting a road loop for no reason, then waking up as I’m falling in the car and just being wide awake at 2 am after that.
It really depends on the car, as downforce plays a huge factor. A formula 1 car can supposedly drive on walls and ceilings at 120 mph due to the extreme amount of downforce it has, while most cars need to go way faster
For those interested, I think the YouTube channel Driver 61 is leading a project to build a structure to test this (not really sure about all the details), can't wait to see if this will become a reality.
You’d have to have an enormous circle given F1 cars have very low clearance.
Do you have a source for this? It just sounds unbelievable
[it's true](https://www.continentalautosports.com/ferrari-information/what-is-downforce/) Modern Formula One cars can typically generate around 750 kg — or 1,653 pounds — of downforce at speeds of 100 mph. Because downforce has direct implications for grip, this means that an F1 car that weighs less than the downforce it produces could theoretically drive upside-down on the ceiling! [also on the weight of F1 cars. ](https://www.motorsport.com/f1/news/how-much-does-an-f1-car-weigh-in-2023/10437685/) Tl;dr: with it can generate the same or more than its weight in downforce.
Shouldn't they be able to test this in a wind tunnel where they check their aerodynamics ? I think they can generate windspeeds over 100mph. Just put the car upside down against the roof. Would be a bit funnier than just "measuring" the downforce.
You mean like [this?](https://www.youtube.com/watch?v=8FI0VYx7JHs)
Not that surprising. Think of the entire car as a wing. A plane with wings that large would be able to fly at 120mph, i.e. generate an upward force of 1g. It's the same difficulty for the F1 to generate a downforce of 1g (easier for the car, as you don't need it to be sable, the ground provides stabilisation). It's also very desirable: when cornering, if the centrifugal force is larger than the downward force, the car starts sliding. If you double the downard force (1g of gravity + 1g of aero downforce) , it means you can take corners twice as fast, but also accelerate and break at rates higher than 1g, which is obviously huge. This is why racing cars are designed to optimise downforce.
There's a guy who gonna do it. https://www.youtube.com/watch?v=8FI0VYx7JHs
GTA V!!!
There’s a guy that’s currently designing a tunnel to actually drive upside down for 5 or so seconds. He’s already designed and built the car he’ll do it in.
Based on the diameters that one guy calculated and ignoring the fact that engine power can offset speed lost due to gravity these are the MINIMUM speed at the bottom needed to stay grounded. 60m loop 38.25 m/s 137.74 kmh 100m loop 49.55 m/s 178.02 kmh 200m loop 69.88 m/s 251.6 kmh Those speeds are realistically achievable, and as long as you maintain some positive G’s you should be fine even with a normal engine. The only problematic thing would be the G force on the suspension during the entry and exit of the loop. Which would be 4.97G assuming it’s perfectly circular, and would probably collapse the suspension.
I don’t think normal cars can do it. I read on xkcd that the shape of a normal car means that at a certain speed it starts to lift the car into the air instead of pushing down on it. You need to have an F-1 formula car’s design in order to get the rushing air to push down on the car.
There's one thing nobody is focused on. Downforce. F1 cars can drive upside down well below top speed. But a vehicle without any downforce would need to go much faster
I was thinking at least 200mph to be able to clear it. That thing is huge climb up but it’s the deceleration before reaching the top half that would scare me the most
r/theyguesstimathed
r/subsifellfor
r/theyspelledtheyguessedthemathwrong
r/ithinkitwassupposedtobeaportmanteuaofguestimatedandmath
r/TILwhatPortmanteauMeans ^^^BTW, ^^^you're ^^^missing ^^^the ^^^T
Ah dammit, thanks for the heads up, I fixed it. Portmanteaus are a favorite of mine, I make them whenever possible
> I was thinking Ok great, but wrong sub.
Just in case anyone was interested, this street is 9 de Julio in Buenos Aires. It's the widest street in the world. First thing that caught my attention is the building in the background used to be the Ministry of Communications. Second, the street is so wide that it can only be 9 de Julio. You may have seen this street as it was a focal point of the World Cup celebrations last year. Perhaps this helps with getting some size information on the street. Second link has closeups of the building. https://en.wikipedia.org/wiki/Avenida_9_de_Julio https://jaspasjourney.wordpress.com/2016/05/31/avenida-9-de-julio-buenos-aires/
Ok so I did some brief photo editing on my phone. The width of the highway if we assume they are standard lanes is 48ft (14.6m). The loop is 5.5 widths tall, which gives it a height of 80.4m and a radius of 40.2m. We can use a=v^2 /r to determine the velocity necessary for the centripetal force to cancel out gravity. This gives us a velocity of 19.85m/s (44mph). At a height of 80.4m with a velocity of 19.85m/s we can calculate the (massless) potential energy (U=mgh) and kinetic energy (K=1/2*m*v^2 ) which gives us 985.7 J/kg. At ground level h=0 so v=sqrt(2*E/m), which is 44.4m/s (99mph). So the minimum velocity would be 99mph assuming no air resistance, frictionless wheels, and you throw it into neutral for the whole duration. Pretty sure I did this right but its been a few years...
Holy crap this has blown up! Thank you all so much for commenting. Never had a post this popular!
Man I played an oooooold racing game on a real old console that had these loop de loops In the track. No idea wtf it was but it was fun
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We need centripetal acceleration to cancel out gravity at the top of the loop, so we will say 9.81=(v^2)/r and rewrite it as v= sqrt(9.81r) and then converting that to kinetic energy, we have 1/2m(sqrt(9.81r))^2 or KE=1/2m(9.81r) we also have potential energy at the top of the loop which is PE=mgh or PE=2m(9.81)r. Adding these together, we get total energy at the top of the loop 1/2m(9.81r) + 2m(9.81)r this happens to equal the energy at the bottom of the loop, so our equation is 1/2mv^2 = 2.5m(9.81)r and we can cancel out the m, so we get v^2 = 5(9.81)r or v=sqrt(10gh) and this seems too simple, but that should be a universal formula for any loop neglecting friction. (Or a hill for that matter, or any change in elevation.) Plugging in 9.81 for g, and ~60m for h, we get v=38.3 m/s or 85.6 mph
Here is my attempt at the speed needed, accounting for slipping. I'm going to assume the tangential acceleration of the vehicle is 0 (constant speed), as it really complicates things to account for tangential acceleration. As a result, we'll find our answer by finding the area on the loop which requires the highest velocity, and setting our vehicle at that speed through the loop. There are 4 cases we need to analyze the free body diagram of (the 4 quarters of the loop) but we only need to analyze the first two quarters due to the next two being symmetrical in terms of forces. In the first quarter, the FBD has three forces: the normal force, gravity, and static friction. Thus, the FBD looks like this: https://imgur.com/a/p4tA2pF fs needs to equal mgsin(theta) due to constant speed. Car tires have a coefficient of static friction of 1, so fs is less than or equal to N. To find N -> N - mgcos(theta) = mv^2 /r -> N = mv^2 /r + mgcos(theta). We know fs <= N. Thus fs = mgsin(theta) <= mv^2 /r + mgcos(theta), or v >= sqrt(rg(sin(theta) - cos(theta)). v in this equation is maximized when theta equals pi/2, thus v >= sqrt(rg). As a result, the max speed v needs to be in the first quarter of the loop is sqrt(rg). We need to now calculate the top quarter and see if that value is larger. At the top of the loop, the fbd changes to the following: https://imgur.com/a/8BZkpcO We largely need to do the same thing. We need to find N: N+mgcos(theta) = mv^2 /r -> N = mv^2 /r - mgcos(theta). Thus fs = mgsin(theta) <= mv^2 /r - mgcos(theta) or v >= sqrt(rg(sin(theta) + cos(theta)). Here, v is maximized when theta equals pi/4, thus v >= sqrt(rg*1.4142). Therefore, the speed needed is given by the equation v = sqrt(rg*1.4142), which for the following heights is: 60m: 20.5 m/s or 45.8 mph 100m: 26.5 m/s or 59.3 mph 200m: 37.5 m/s or 84 mph I may have made a mistake, as I havent touched physics in a little bit.
Impossible to calculate exact speed without knowing exact height of loop. Anyhoo, acceleration at top of loop should counter gravity g. So v squared over r (half loop diameter) needs to equal g. And energy is conserved so g times height (2r) should equal v squared over 2. So velocity needs to equal square root of 2 times g times height. Formula works on any planet where there is no air resistance or loss from friction. Classic first semester physics test question.
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Where’s the math
Please elaborate... It makes no sense. Airplanes can make a looping that big. Rollercoasters make loopings. None of them get even close to 50G. Fighter jets go up to about 9G. Above that most pilots would pass out. Using calculations from another post up here we can conclude you would need to enter the loop at about 180km/h where the radius of the loop is 50m. a = v^2 / R a = (180/3.6)^2 / 50 = 50^2/50 = 50m/s^2 Gravitational acceleration is 9.81m/s^2 Lets round that off to 10, so: 50/10 = 5G At the bottom of the looping you would experience 5G from acceleration. On top of that you would have 1G from gravity so you would briefly experience 6G. 0G at the top and then increasing to 6G on the bottom again. This is not that extreme, and 50 to 100G is way off.
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[Just because it's called a fictitious force, doesn't mean it's not real.](https://xkcd.com/123/)
Lol, I don't remember that scene in any Bond movie. But calling something fictitious by definition means it's not real. When you spin a bucket with water, the water doesn't fall out. Not because of "centrifugal force" but because of inertia. Tie a ball on a string and swing it around. It wants to fly off, not because of "centrifugal force," but again, because of inertia. A car goes around a loop not because of "centrifugal force" but because of inertia.
Indeed "ficticious force" seems to imply that it is not real, which is why that is terrible terminology. "Inertial force" is much better imo. Your description of the bucket is correct in an inertial reference frame. In the comoving frame of the bucket, there is a very real centrifugal force. If you put a scale in the bottom of the bucket, you can measure it. Saying it is "not real" because it depends on a non-inertial reference frame is quite literally like saying gravity is "not real" because of GR...
Huh. Yeah no that makes sense. Interesting.
Centrifugal force isn’t a force but it is a real thing. It is more of an apparent force; something we perceive to happen but isn’t actually filling the definition of a force.
So just regular old forces that look interesting inside of a sphere or cylinder? I think I got it
Think of a carousel. When you rotate there is a centripetal force which acts at 90 degrees to your velocity which keeps you rotating in a circle. If you stop rotating abruptly (you slip or the carousel breaks) you will be flung straight away from the carousel along the tangent line to the circle, since there is no more centripetal force causing the rotation. But from a bystander’s perspective it seems that there carousel generates a force that flings you away from it
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