# How fast would an object have to be moving to

1. Dec 3, 2011

### CuriousStrang

move up a vertical surface with no interruption?

I've seen this represented so many times in various fictional media, that I'm curious if there is some real life phenomena from which it draws inspiration?

There are probably factors such as the size of object involved surely?

Anyway, this is the first of hopefully many posts here from me. So Hello. :)

Hopefully this is the appropriate forum.

edit: And this is of course, assuming that the object doesn't have some adaptation to aid it, such as with arthropods.

Last edited: Dec 3, 2011
2. Dec 3, 2011

### Curl

I don't understand what you are asking.

3. Dec 3, 2011

### CuriousStrang

Sorry, how fast would an object(let's say a human like object for simplicities' sake), theoretically have to be to overcome gravity and run up a vertical surface like a wall?

4. Dec 3, 2011

### Curl

Any speed works.

Depending on how far up the wall you want to go, you'll need more initial speed.

5. Dec 3, 2011

### CuriousStrang

Suppose, its a 100 foot wall?

Last edited: Dec 3, 2011
6. Dec 4, 2011

### DaveC426913

No, you need to apply a force on the wall to get the traction to be able to move up it. This will act to push you away from the wall. If you lose contact with the wall before you reach the top, you'll pitch over and down.

It is possible in principle to rise a certain amount up a wall but the key is that you need to reach the top before your forward (horizontal) velocity reaches zero. That's going to happen very quickly. (You are essentially decelerating from your top horizontal speed to zero over a distance shorter than the length of your legs.)

7. Dec 4, 2011

### Ken G

We need more information. Curl is right that any speed works, you just go farther up the wall, but then you also don't need any contact with the wall-- it's just a trajectory. The question seems to be implying that traction needs to be retained with the wall, and that moving faster somehow increases the traction. I think the model being visualized must be like how a dancer runs at a wall in order to be able to run up it and flip over. In that case, the dancer needs speed to help store potential energy in their muscles and bones, which helps them provide the energy to get up the wall, and it also helps increase the "normal force", which is the contact force the wall provides to keep the dancer out of the wall.

The reason the normal force is important is that the maximum force of static friction is often treated as proportional to that normal force. So the dancer wants to maintain a strong frictional contact with the wall while the wall is decelerating the dancer's horizontal motion, and meanwhile the dancer is using their muscles to drive themself up the wall, taking advantage of that maximum possible static friction force. If this is the kind of scenario the original poster has in mind, then it is a more detailed calculation requiring knowledge of the coefficient of static friction with the wall, and you might need other things like the mass of the object, etc.

8. Dec 4, 2011

### CuriousStrang

I am indeed sort of thinking of it from a visual of a person running up a wall.

But more specifically the scale of the question I'm visualizing is rather impossible, in which the person can somehow keep going without ever completely losing their horizontal acceleration--or at least for a reasonably (yet still impossibly) long height, which is why I settled on the 100 foot wall.

I suppose we can just assume the person is of average size, say 5'9" and 150 lbs? And the wall could be something fairly uniform like concrete?

9. Dec 4, 2011

### DaveC426913

It is perhaps worth pointing out that, if provided two walls at 90 degrees, Jackie Chan performs this feat quite nicely, going up at last ten feet.

In theory, if the person's strength and stamina are not limited, they could climb this way to any height.

10. Dec 4, 2011

### zoobyshoe

As KenG says it's just a trajectory, the person won't have any traction on the wall. The question becomes how much vertical speed an object would need when leaving the ground to make it to a height of 100 feet before gravity brings it to a stop.

The mass of the object is immaterial here: all objects fall at the same acceleration, therefore they all are decelerated by gravity at the same rate when impelled upward at any given initial velocity.

The initial upward speed any object would need, therefore, to reach a height of 100 feet is 54.8 mph. (That is simply the same speed they'd have when they hit the ground if they fell from 100 feet starting at 0 velocity. All you need to do is reverse it.)

No one can run that fast, obviously, and even if they could, I think they might well be killed in the attempt to suddenly change their trajectory at the bottom of the wall.

11. Dec 4, 2011

### DaveC426913

You're discounting the traction he could get while climbing. That's the whole point of the feat.

12. Dec 4, 2011

### zoobyshoe

How could he get any traction? There's no force pushing him against the wall. The feat, as he described it, isn't possible.

The Jackie Chan thing you described has very different conditions. To the extent his shoes don't slip, he can treat the two perpendicular walls nearly as if they were parallel and play one off against the other. If the acrobat in the OP scenario pushes off against his wall, he has no other wall to prevent his continued motion in the new direction.

13. Dec 4, 2011

### rcgldr

If the coefficient of friction is enough, then the normal force related to lateral accelerations could allow a vertical force greater than his weight. For a pair of perpendicular walls, it would seem the coefficient of friction would need to be greater than 1. Jumping up between a pair of parallel walls would be easier (this has been done or simulated in some movies).

14. Dec 4, 2011

### zoobyshoe

I'm talking about the case of a guy running straight up a vertical 100 foot wall.

15. Dec 4, 2011

### DaveC426913

16. Dec 5, 2011

### zoobyshoe

No, I'm talking about a guy running directly toward a wall, then trying to change his trajectory suddenly to a vertical one at the bottom of the wall.

What I'm saying is that the speed required to have enough momentum to rise to a height of 100 feet would 1.) kill or seriously injure the acrobat in the attempt to use the deceleration represented by an encounter with a wall for traction up the wall, or 2.) if he non-injuriously converted it (somehow, I don't know how), he would have to do so at the expense of all further traction: he would just launch himself vertically and grab the top of the building at the top of his trajectory.

I think the real life limit on this sort of thing is the case KenG mentioned of dancers who use a wall to flip themselves over backward. This is probably a natural consequence of trying to run up a wall while standing erect against the wall's normal force: you'd start to rotate backward over your CoG and have to go with that to land on your feet. I've never tried it, though.

17. Dec 5, 2011

### CuriousStrang

Dave your assumption was more or less correct, that is the visual that I am imagining.

18. Dec 5, 2011

### Staff: Mentor

You know the two-step backflip dancers do against a wall? That's the limit.

Did you see what Neo did in his first fight scene in the Matrix (maybe a 5-step backflip)? Not possible.

19. Dec 5, 2011

### Ken G

The problem is, to maintain traction, one needs a horizontal normal force, if all we are talking about is static friction (obviously free climbers can "run" up a cliff of any height if they can avail themselves of little nooks and crannies that can generate a vertical force without any horizontal normal force). That horizontal normal force will produce horizontal acceleration, so the initial horizontal speed sets a kinematic limit on how much time the person can remain in contact with the wall. If the coefficient of static friction is less than 1, as for usual surfaces, then the horizontal force must exceed the person's weight. That means their horizontal acceleration must exceed g to get sufficient traction, and the more it exceeds g, the more traction they can get and the faster they can accelerate up the wall. If we say they are not trying to accelerate up the wall, but are just moving at a constant speed up the wall, then the horizontal acceleration only needs to be a little more than g. That means the initial speed needs to be at least (roughly) the square root of 2gL, where L is the length of their legs. For legs of 1 m, that means about 4-5 m/s. The faster you can go than that, the more upward push you can get against the wall. The time it will last is v/a ~ 2L/v, and the acceleration is ~v2/2L, so the distance of the wall is about at2/2 ~ v2/2L*(2L/v)2 ~ 2L. So it will be hard to get more than about 2 meters up the wall, regardless of how fast you can run-- in agreement with what dancers do. My calculation is too rough to rule out Jackie Chan's 10 feet, but I wouldn't believe any higher.