Is moving so fast you leave your shadow behind a calcable feat?

[MasqueTLDF]

If so, what tier would it be and is there an accepted calculation for a feat of this sort? Two examples.

 

[DontTalkDT]

Depends. For flying characters it works. For characters that touch the ground the feat makes no sense as, regardless of speed, you would never leave your shadow behind if you make direct contact with it (i.e. the ground it is on).

 

[MasqueTLDF]

Depends. For flying characters it works. For characters that touch the ground the feat makes no sense as, regardless of speed, you would never leave your shadow behind if you make direct contact with it (i.e. the ground it is on).

if the character performing the feat is in the air when it happens what tier would the feat be?

 

[SeijiSetto]

For characters that touch the ground the feat makes no sense as, regardless of speed, you would never leave your shadow behind if you make direct contact with it (i.e. the ground it is on).

is it not possible to simply travel faster than the light that would hit the ground to cast your shadow

 

[TheGreatJedi13]

toon force if done in a gag manner

 

[MasqueTLDF]

toon force if done in a gag manner

If not?

 

[DontTalkDT]

is it not possible to simply travel faster than the light that would hit the ground to cast your shadow

While walking? Nah.

You have a shadow below your shoe that is touching the ground. How long did that shadow need to reach the ground? Well, the distance is 0 meters, speed of light in air is 299,702,547 m/s, therefore the time it took is 0 m / 299,702,547 m/s = 0 s. No matter how fast you move you are not leaving that behind.

And now you might say that the shadow below your shoe doesn't count, but the shadow that is cast from the part of your body that is 0.00000000000000000001m away from the ground also just takes 0.0000000000000001 / 299,702,547 = 0.0000000000000000000000003337 s.
In general, you will for any ε > 0 find some distance x > 0 so that the parts of your body that are at most x m away from the ground will cast a shadow on the ground in less than ε seconds. The human body is continuous so from the 0s time from below your feet to the time for light from your head to reach the ground all values are given.

That's why you would need to fly so that you are some distance from the ground. Then you can calculate (and the position of the observer is relevant here) how fast you would need to move so that your are some distance away from your shadow in the view of the observer by the time the light of the shadow first hits the ground and then hits their eyes.

While walking all that happens would be the shadow stretching really long.

 

[MasqueTLDF]

This is really complicated lol, how could i format this into a calc request? Can I use your statement as reference for the formula?

 

[DivinumAeterna]

If not?

Anywhere from MFTL to infinite (based on my understanding of physics) depending on if the shadow moved even a little or not at all in the frame that the person traveled, especially since in your example there's quite a considerable distance between the character and the spot they were standing in (meaning you can dismiss this as an outlier due to author having fun drawing cool-looking panels, if nothing else came close to this feat or if this implication of speed completely breaks the narrative of the work in question imo)

 

[DontTalkDT]

This is really complicated lol, how could i format this into a calc request? Can I use your statement as reference for the formula?

That... isn't even the formula. That actual calculation is more complicated (and doesn't work for walking characters, so I don't think it works for your feats).

Let's see... in general the character is x meters from the ground (and stays at that distance). The shadow is at location S, the observer at location O and the character is seen at location C while its shadow is at S. Furthermore let S' be the location above the shadow at which the character was when they cast the shadow (for simplicity sake we will assume the light comes from straight up above).
The distance light covered for the shadow to be visible to the observer after the character was at S' is given by the distance needed for it to reach the ground and then the observer. So d(S',S) + d(S,O). (Where d is of course the metric)
The time of the crossing is that, divided by the speed of light in air. So (d(S',S) + d(S,O)) m / 299,702,547 m/s := t1.

The distance covered for the light of the character to reach the observer is meanwhile d(O,C). The time that light needed to reach their eyes is d(O,C) m / 299,702,547 m/s := t2.
Then t1 - t2 is the amount of time the character had to move from S' to C, as t1 second ago the character must have been at S', yet t2 second ago they need to have been at C.
So the total speed is d(S',C)/(t1-t2).
...I think. Just scribbled that down real fast.

All of this does not consider possible relativistic effects, like relativity of simultaneousness. (if relevant... idk)



Feel free to use that information as you like.

 

[MasqueTLDF]

Anywhere from MFTL to infinite (based on my understanding of physics) depending on if the shadow moved even a little or not at all in the frame that the person traveled, especially since in your example there's quite a considerable distance between the character and the spot they were standing in

hm...

(meaning you can dismiss this as an outlier due to author having fun drawing cool-looking panels, if nothing else came close to this feat or if this implication of speed completely breaks the narrative of the work in question imo)

The feat's performed twice by different characters (their feats are even compared to each other in the series) and it's implied that most mid tiers are capable of the same thing. Doesn't really seem outlier-ish to me.

That... isn't even the formula.

Like I said, it's complicated. I have an IQ of 12 bro, i dont understand nun of this.

That actual calculation is more complicated (and doesn't work for walking characters, so I don't think it works for your feats).

The characters in the feats weren't walking or even on the ground when the feats were performed though? Yuuto dashed toward Issei and Bennia is just always levitating.

Let's see... in general the character is x meters from the ground (and stays at that distance). The shadow is at location S, the observer at location O and the character is seen at location C while its shadow is at S. Furthermore let S' be the location above the shadow at which the character was when they cast the shadow (for simplicity sake we will assume the light comes from straight up above).
The distance light covered for the shadow to be visible to the observer after the character was at S' is given by the distance needed for it to reach the ground and then the observer. So d(S',S) + d(S,O). (Where d is of course the metric)
The time of the crossing is that, divided by the speed of light in air. So (d(S',S) + d(S,O)) m / 299,702,547 m/s := t1.

The distance covered for the light of the character to reach the observer is meanwhile d(O,C). The time that light needed to reach their eyes is d(O,C) m / 299,702,547 m/s := t2.
Then t1 - t2 is the amount of time the character had to move from S' to C, as t1 second ago the character must have been at S', yet t2 second ago they need to have been at C.
So the total speed is d(S',C)/(t1-t2).
...I think. Just scribbled that down real fast.

All of this does not consider possible relativistic effects, like relativity of simultaneousness. (if relevant... idk)



Feel free to use that information as you like.

Uh, ok! Thanks a ton!