I thought this might’ve been asked before, but I searched and couldn’t find it.
I am always willing to suspend disbelief for a good story, but I sometimes find myself wondering:
Just how much pull/force is exerted at the anchor point of Spidey’s webline while he swings through the air?
I am not an engineer, but figured many a Doper were and could answer this. My hypothesis is that so much force would be exerted IRL that 90% of the time, the webline would pull off the building, leaving Spiderman hanging (well, then falling).
I guess some assumptions would be:
Spiderman weighs, what 160 - 175lbs. (PParker is supposed to be small, right?)
Avg webline - what? 30’
He is moving at what? 30 - 40mph as the line affixes to a building?
If you take a crack at calculating this - thank you. If you change or add assumptions, please state what they are.
As a follow up - what kind of force would it take to dislodge most superficial pieces on a building - e.g., pull a window out, yank off a piece of masonry or something like that?
Thanks!
PS: I thought this should be in GQ since it asks for a math answer. If the mods think Cafe Society - please move it.
Assumption: Spider-Man has a mass of 75 kg (~165lb).
Assumption: Spider-Man’s swinging web is cylindrical (circular in cross section) having an unstressed length of 10m, a diameter of 2mm (!), and a cross-sectional area of 3.14 * 10[sup]-6[/sup] m[sup]2[/sup]
Assumption: For swinging purposes, Spider-Man uses a silk similar to the black widow’s dragline silk. In this article the authors derive a Young’s Modulus of 11 GPa.
Young’s Modulus is defined as E = (length / change in length) * (force / area), and the spring constant is defined as k = (force / change in length), so we can simplify Young’s Modulus to be E = k * L / A. We’ve assumed L and A above, so we have
11 GPa = k * 10m / 3.14E-6
k = 3454 N / m
So there’s your spring constant for spider dragline silk in this case. I’m pretty rusty on my dynamics, but someone should be able to take my derived values and model this as a spring pendulum for a heap of extra credit. For the simple case, however, it’s enough to note that his cable isn’t going to stretch too much to ruin a “constant length” assumption (we hope) and so we can probably use the equations found here to calculate the tension (and therefore the force on the masonry).
I’ve got some other work to do but I’ll return to finish this up when I’m done.
I’ve done the calculations before, and assuming that he starts the swing with the web horizontal, his own height is negligible compared to the length of the web, the web does not stretch, and air resistance is negligible, he’ll be feeling three gees at the bottom. Which means that the force at the anchor point would be three times his weight. If we relax any of the simplifying assumptions above, this force decreases.
Thanks Jurph and Chronos. Chronos, based on your simplified ballpark estimate, that would show a force at the anchor point of about 450 - 600lbs (assuming he weighs 150 - 200lbs).
Next question - is that a lot? How do I put that in context compared to other common “pull” situations - playing crack the whip, say, or a truck pulling a boat. Am I correct in thinking that Spidey is likely to “yank himself off” (get your head out of the gutter!) a number of his anchor points? Or is this amount of pull actually not all that bad?
As a side note, Sci-Am determined that the apparent thickness of the web-fluid he shoots out is plenty thick enough to hold him up. Spider silk is strong stuff.
It’s exactly as much pull as you’d get from three guys on a rope just hanging there, without any jerks. And I think that Spider-man’s webs typically have a fairly large attachment point (several square inches), so I’m not seeing any problem, as long as the adhesion is strong enough (that being of course uncalculable, without knowing the properties of Spidey-web).
For the followup question in the OP, keep in mind he does not swing with the web vertically oriented, but possibly at a 30deg angle to vertical, so only the component of force normal to the wall (horizontal component) would be contributing to any bricks coming loose or whatnot. Building do quite well at withstanding the vertical (downward) component of this force. With a 30-60-90 arrangment, the horizontal component is half the full force (the hypotenuse of the triangle).
This question has indeed come up before. You might not have been able to find it because the OP of the earlier thread, entitled A question about g-forces and Spiderman’s supersphincter, asked specifically about the possibility of the Spidey-colon being accelerated out of the Spidey-anus due to extreme g-forces. An interesting spin on the question, but perhaps making it hard to recognize with a search.
Or it could have been that you didn’t search in Cafe Society. Either way.
Again, this is all assuming he always fastens his web onto a solid piece of wall and not, say, a window, air conditioner, passing bird, or the face of somebody who’s looking out the window to see what all the commotion’s about.
Funny how that sort of thing never seems to happen…
Let’s do some back o’ the envelope calcs here, and base our assumptions on what we see in the movies.
Spidey is about 80kg. Sure he’s a small guy but he’s packing a lot of muscle.
We see Spidey overtaking traffic on city streets, so let’s say he’s moving about 45mph (a little over 20 meters per second) parallel to the street.
We also see that his web is fixed somewhere up above to the side of a convenient building. Let’s pick 10 stories up at 12 feet per story, that’s about 36.6 meters.
Basic physics gives us the centripetal acceleration = v^2/r.
a = (20^2)/36.6 (meters per second squared)
= 10.9 m/s^2.
No problem there, that’s only a little more than 1G. Force required to hold the line at the other end is only about 10% over Peter’s body weight.
If we use your example with a much shorter strand (10 meters), acceleration climbs to about 40m/s^2 (4G). Force on the attachment point is now up to 3200N or about 700 pounds. I’d be a little more careful about where I stuck my web if I was him - wouldn’t want to stick it to a window pane.
Strength of the web shouldn’t be much of a problem. The 825 pound microline on my parachute is fairly thin stuff so Spidey’s web is doubtless well up to the task (especially since we know he can dangle cars and stuff from it).
My biggest concern would be getting severe line burn from holding on to the stuff.
