Which Spider-Man Is Stronger: Tobey Maguire or Tom Holland?
Spider-Man began as a comics character, he has actually made his method to live-action video numerous times. I keep in mind seeing him appear on The Electric Company in the 1970s for a brief act; it was cool however a little odd. In the modern-day period of live-action Spider-Man films, we had the Tobey Maguire variation, followed by Andrew Garfield’s turn, and lastly the Tom Holland variation that appears in the existing Marvel Cinematic Universe. We got an opportunity to see all 3 in Spider-Man: No Way Home, which was fantastic, plus an excellent reason to respond to the concern of whether MJ might truly hold on throughout among Spidey’s swings.
Now it is time to ask an even harder concern: Which variation of Spider-Man is the greatest? Let’s compare the Maguire variation in 2004’s Spider-Man 2 to the Holland variation in 2017’s Spider-Man: Homecominggiven that they carry out comparable actions: a test of strength that includes utilizing Spidey’s webs to limit a moving lorry. Maguire’s Spider-Man stops a runaway subway train, and Holland’s usages webs to hold a splitting ferryboat together. (It would have been terrific to consist of Garfield’s variation in this contrast, however there’s simply not a scene that reveals a comparable accomplishment of strength.)
Stopping a Subway Train
Here’s the circumstance in Maguire’s Spider-Man 2, which you can enjoy in this clip: After a fight with a bad man, Spider-Man discovers himself at the front of an out-of-control subway train. There are a lot of individuals on the train, so he requires to conserve them. He tries to slow the train by jamming his feet down onto the track, however that does not work. He shoots some webs at the structures on both sides of the track and holds on. The webs stretch and– spoiler alert– the strategy works. Spidey stops the train.
If we approximate the force needed to stop this train, that will likewise be a quote of Maguire’s strength.
Let’s begin with some physics ideas. Expect I have a things with a mass (m) moving with a speed (v). If you use a force to this item, it will experience a velocity based upon Newton’s 2nd law, which mentions that the net force amounts to the item of its mass and velocity (Fnet = m × a). In this case, that object is a train, and the force is the backwards-pushing force from Spider-Man’s webs, which he is keeping.
If I approximate the mass of the train and discover the velocity, I can compute that force. We specify velocity as the rate of modification of speed. As a formula, it appears like this:
For this train, the last speed (v2would be absolutely no, given that it stops. That implies I will require both the preliminary speed (v1and the stopping time from when Spider-Man shoots the webs till it in fact stops (Δt). The time is quite simple, due to the fact that I can get that from the video– it takes 36 seconds.
What about the preliminary speed? OK, we are going to require to do a little work here. In the scene, we see a good view of the movement of the train right when Spider-Man shoots his webs onto the structures. I can drop this clip into Tracker Video Analysis and get position-time information from the area of the train in each frame. (I utilized a brief sector of the train’s movement prior to Spidey’s webs get hold.)
There is another small information I require to get the proper range scale, which is the length of each train cars and truck. Spider-Man 2 occurs in New York City, this scene was in fact shot in Chicago on the L train, which is brief for “raised.” Those train vehicles have a length of 14.7 meters, precisely the info I require. With that, I get the following information for the position of the train:
Considering that the speed is the rate of modification of position, the slope of a position-time chart would be the speed. This offers the train a beginning speed of 27 meters per 2nd, or 60 miles per hour. That thing is going quite quickly, however L train cars and trucks in fact have a leading speed of 70 miles per hour. Still, it’s a good idea Spidey exists to stop it. Utilizing that worth for the preliminary speed and a time of 36 seconds provides a velocity of 0.75 meters per 2nd per second.
Next, I require the mass. Each vehicle has an empty mass of 26,000 kgs and a capability of 34 seats, or 123 overall travelers. From the clip, the train looks quite crowded, however not at complete capability. Let’s simply state there are 60 individuals per automobile, and each rider has a typical mass of 70 kgs.
I require the overall number of train automobiles. We never ever get a best shot, however I’m going to think there are 5 vehicles. That puts the overall mass at 151,000 kgs. Factoring in this mass and the velocity, we get a stopping force of 113,000 newtons.
That’s the overall force pulling on the train to slow it down. Keep in mind, there are webs on both sides of the train. Because Spider-Man keeps webs on both sides, he’s basically simply applying a force equivalent to the stress in the web. That suggests that he’s just applying a force that’s half the overall worth of the stopping force.
The very same thing occurs when you pass a string over a wheel: It enables you to double your pulling force. (Spidey is basically serving as the wheel in the sheave system.) Spidey’s applied force would be half of the 113,000 newtons, or 56,500 newtons. If you wish to transform that to royal systems, it would be 12,700 pounds. That’s like holding up a male African elephant. You can see my estimations here.
I’m going to put the Tobey Maguire Spidey strength at a worth of a minimum of 56,500 newtons. Obviously, we do not understand if that’s his optimum limitation, however it’s at least a beginning worth. I have to admit that I may have cheated. I presumed that the force that Spider-Man applies on the train is consistent. That’s most likely not the case. If spiderwebs resemble elastic band, bungee cables, and a lot of elastic things, the more you extend them, the higher the force it requires to hang on to or pull them. Given that I do not have any proof that these webs act like bungee cables, I’m simply going to stick with my constant-force evaluation.
Holding a Splitting Ferry Together
Now we can relocate to Holland’s variation in Spider-Man: Homecoming, which you can see in this clip. After an alien weapon goes off and divides the Staten Island Ferry right down the middle, Spider-Man does his finest to spot the boat up with his webs to avoid it from sinking. It gradually begins to separate, with the 2 halves breaking lengthwise as some of the webs holding them snap apart. In a desperate relocation, he gets webs connected to both sides and attempts to pull the ferryboat back together. I’ve drawn a sketch of the vital part (not to scale).
Do you see what we have here? Similar to with the stopping of the train, Spider-Man is hanging on to webs and most likely applying a big force. It’s an ideal contrast. Now we simply require to discover a worth for that force.
This computation is going to be a bit more tough from a physics viewpoint. There are numerous aspects that we simply can’t put a worth on. Just how much of each side of the ferryboat is filled with water? The number of individuals are onboard? What about the webs that didn’t snap and are still assisting hold the boat together?
It’s great. When a physicist gets an unsolvable issue like this, we simply turn it into an easier issue. Yes, it appears like unfaithful, however a minimum of it will provide us a ballpark response for Spider-Man’s strength.
Here’s my easier issue. A block rests on the edge of a table and is slanted over the side at an angle. A horizontal string is connected to the block, and somebody pulls on it to avoid the block from tipping over. This is sort of like what’s taking place to each half of the splitting boat, with each side tilting away from the. (I’ve presumed that the bottom of each half of the ferryboat is still linked to the other half. Otherwise, both parts would simply sink directly down rather of toppling.) The block being on the edge of the table mimics this, due to the fact that it’s leaning at a comparable angle and held up by a string, simply as Spider-Man and his webs are the stress holding the 2 boat halves together. Here is what it appears like:
Notification that there are 3 forces acting upon this tipping block. The very first is the force that Spider-Man applies on the block, which is identified FsNext we have the downward-pulling gravitational force (Fg. The force from the edge of the table (FT.
If this block is at rest, then 2 things should hold true. The net force needs to amount to absolutely no (so that it does not speed up), and the net torque needs to be absolutely no (so it does not alter rotational movement). Simply a fast suggestion about torque: You can consider this as a “rotational force,” or a force that triggers modifications in rotational movement. The magnitude of a torque (τ) depends upon the strength of the force (F), the range from the force to a pivot point (r), and the angle of the force (θ).
In our tipping block issue, we just require to handle 2 torques: the one due to Spider-Man’s pull and one from the gravitational force. The force applied at the pivot point is pressing At the pointso that it has a range of r = 0 and absolutely no torque. That indicates that the other torques should be equivalent and opposite in order to keep the half of the ferryboat fixed.
I can utilize this formula to discover the unidentified worth of the force that Spider-Man applies. I’m going to require some worths– like the mass of half a ferryboat and the length of the entire thing. Wikipedia does a terrific task of noting statistics that you would not believe you would ever in fact require, so I understand the Staten Island ferryboat has a mass of 3,335 gross lots (3.38 million kgs) and a length of 94 meters. I will likewise require the tilt angle of the boat, which I managed determining the angle onscreen: 20 degrees.
When I plug in all my worths, plus some other price quotes, I get a force from Spider-Man equivalent to 12.7 million newtons, or 2.8 million pounds. Now we can make a contrast: The Tobey Maguire Spider-Man pulls with a force equivalent to the weight of one elephant, however Tom Holland’s pulls with an equivalent of 219 elephants.
Keep in mind, this worth is based upon estimates, and a few of these evaluations are most definitely incorrect. It’s extremely possible that the force required to hold up the ferryboat might be half my determined worth and the train-stopping force might be two times my worth. Even then, we are talking about 2 elephants compared to 100. Tom Holland’s Spider-Man is absolutely more powerful.