help with math and quantifying time dilation

[math_noob_1]

hello, this is my first post here! and i hope it to be the first of many

i have a question about time dilation or at least something similar to that

if someone shoots a gun with a muzzle velocity of 910 m/s at a target 5 meters away - it should take 0.005 seconds for the projectile to reach the target

now lets say a 3rd party observer in the room sees the bullet in real time slow motion - and the bullet takes 2.5 seconds to reach its target instead of 0.005

is quantifying the speed of the observer as simple as dividing both results?

for example: (2.5 / 0.005) = 500 > observer is 500 times faster than the velocity of the projectile?

thanks!

 

[Deleted member 4993]

hello, this is my first post here! and i hope it to be the first of many

i have a question about time dilation or at least something similar to that

if someone shoots a gun with a muzzle velocity of 910 m/s at a target 5 meters away - it should take 0.005 seconds for the projectile to reach the target

now lets say a 3rd party observer in the room sees the bullet in real time slow motion - and the bullet takes 2.5 seconds to reach its target instead of 0.005

is quantifying the speed of the observer as simple as dividing both results?

for example: (2.5 / 0.005) = 500 > observer is 500 times faster than the velocity of the projectile?

thanks!

Suppose a "fire-proof" ant with a tiny pocket-watch lands on the bullet as soon as the bullet leaves the muzzle.

What would that ant observe - how long would it take for the bullet to strike the target (relative to the ant)?

 

[math_noob_1]

Suppose a "fire-proof" ant with a tiny pocket-watch lands on the as soon as the bullet leaves the muzzle.

What would that ant observe - how long would it take for the bullet to strike the target (relative to the ant)?

hello! thanks for the reply

i don't understand your post in regards to a fire proof ant and landing on what? - i am coming from this perspective that two people are in a room - one shooting the gun and one observing

just assume for logical sake (yes i know, ironic) that the person observing is superhuman fast - like think superman or the flash

is the math as simple as dividing both results to yield how many times inherently faster the oberserve must be?

 

[Deleted member 4993]

hello! thanks for the reply

i don't understand your post in regards to a fire proof ant and landing on what? - i am coming from this perspective that two people are in a room - one shooting the gun and one observing

just assume for logical sake (yes i know, ironic) that the person observing is superhuman fast - like think superman or the flash

is the math as simple as dividing both results to yield how many times inherently faster the oberserve must be?

Mistake of omission - It should have been:

Suppose a "fire-proof" ant with a tiny pocket-watch lands on the bullet as soon as the bullet leaves the muzzle.

Do you know the equation of time-change due to relative velocity between two observer?

I think we have an excellent tutor in this forum - topsquark (a.k.a. Dan) is much better qualified to continue this discussion!! Dan - please take over... I know Jomo is watching...

 

[math_noob_1]

Mistake of omission - It should have been:

Do you know the equation of time-change due to relative velocity between two observer?

I think we have an excellent tutor in this forum - topsquark (a.k.a. Dan) is much better qualified to continue this discussion!! Dan - please take over... I know Jomo is watching...

thanks for another quick reply!

please bear with me, just assume my knowledge is that of a laymen

i have googled the time dilation but it is over my head - i believe it is t = t0/(1-v2/c2)1/2? - but i'm not sure how to applies to a velocity that isn't relativistic?

i was trying to quantify how much inherently faster the observer must of been

 

[Cubist]

I think we have an excellent tutor in this forum - topsquark (a.k.a. Dan) is much better qualified to continue this discussion!! Dan - please take over... I know Jomo is watching...

I imagine that @topsquark is probably exhausted from "an aether" recent thread so I'll have a go at this! If I make a mistake then I'll gladly take a time-out in the corner. I'll let the Flash time my exile.

t = t0/(1-v2/c2)1/2?

? eek, let's make that easier to read...

[MATH] t=\frac{t_0}{\sqrt{1-\left(\frac{v}{c}\right)^2}}[/MATH]
? phew . Plugging your numbers in...

[MATH] 2.5=\frac{0.005}{\sqrt{1-\left(\frac{v}{c}\right)^2}} [/MATH]
[MATH] v=c\sqrt{1-\left(\frac{0.005}{2.5}\right)^2}[/MATH]
[MATH] v \approx 0.999997999997999996\times c[/MATH]
...if I'm correct then that's pretty close to the speed of light! If the Flash is going at this speed AND can see the bullet moving during it's whole journey, then I guess he must be moving in a fairly tight circle around the shooter and target. The G forces experienced would be pretty large!

 

[math_noob_1]

wow, incredible reply cubist - thank you for taking the time to help

i fear however i may of misled you by not explaining the scene correctly - may i message you privately?

 

[Cubist]

You're welcome for the help!
I prefer to discuss in the open forum because other helpers can check things. Also, you may get a quicker response from someone (I don't always log in every day).

i fear however i may of misled you by not explaining the scene correctly

Perhaps, rather than relativity, you're thinking of an observer who can perceive the world at a faster rate, like for example a fly's eye can detect movement much quicker than a human eye?

 

[math_noob_1]

You're welcome for the help!
I prefer to discuss in the open forum because other helpers can check things. Also, you may get a quicker response from someone (I don't always log in every day).



Perhaps, rather than relativity, you're thinking of an observer who can perceive the world at a faster rate, like for example a fly's eye can detect movement much quicker than a human eye?

yes, i believe your example of the fly and their perception is correct - much like a high frame rate camera

here is my work so far - googledoc - superman

i can calculate how much faster superman perceives his enviroment, but he also moves at speeds within that perception - and i am unsure of the methodology to calculate that

i have made notes in the google doc - could you perhaps correct me if am wrong, or offer guidance?

 

[Cubist]

OK, so you're analysing a special effect in a film. The effect seems similar to the so called, "bullet time", effects in The Matrix. There's a serious side to this because high speed film can be a very useful tool for calculating the velocity of fast moving objects.

It might help if you think about the following. As the recording camera's speed, or frame rate, increases then when it is later played back at normal frame rate...

• the perceived time for an event to take place increases
• however, the perceived velocity of a moving object decreases

I copied part of your text and the lines where I put a "###" are my suggested corrections/ replacements...

Spoiler: show suggestions

How long would it take 1 round with a muzzle velocity of 905.21 m/s to complete a trajectory of 5.90 m?

### t = d/v (5.90 m / 905.21 m/s) = 0.0065178 s 
###
### It would take 0.0065178 s in real life

...

### Calculating how much faster the real-time scene would be...
###
### (2s 167 ms) / (0.0065178 s) = 2.167 / 0.0065178 = 332.47
###
### The special effects mimic a high speed camera
### that captures footage 332.51 times faster than
### real-time. This has the effect that velocities
### appear 332.47 times SLOWER. Therefore, the
### projectile's velocity on film appears to be
###
### 905.21 / 332.47  =  2.72268 m/s

How long does it actually take Superman to stop the bullets by moving himself?

It takes 0 s 458 ms on screen for Superman to move himself in front of the bullets

Distance covered by Superman - 2.83 m

### Distance covered by projectiles in that same time...
### distance = speed*time
###    = 2.72268 * 0.458
###    = 1.246987 m
###
### The ratio of the two distances covered (in the same time)
### tells us how much faster Superman is than the speeding bullet
### (2.83 / 1.246987) = 2.26947 times faster
###
### Therefore, his real-time speed is...
###   2.26947 * 905.21 = 2054.35 m/s

There's an alternative way to calculate this. Superman's speed as the viewer perceives it on the screen is given by

speed = distance/time
= (2.83 m) / (0 s 458 ms)
= 6.17904 m/s

The real-time velocity would be 332.47 times faster, which equates to 6.17904 * 332.47 = 2054.35 m/s

 

[math_noob_1]

OK, so you're analysing a special effect in a film. The effect seems similar to the so called, "bullet time", effects in The Matrix. There's a serious side to this because high speed film can be a very useful tool for calculating the velocity of fast moving objects.

It might help if you think about the following. As the recording camera's speed, or frame rate, increases then when it is later played back at normal frame rate...

• the perceived time for an event to take place increases
• however, the perceived velocity of a moving object decreases

I copied part of your text and the lines where I put a "###" are my suggested corrections/ replacements...

Spoiler: show suggestions

How long would it take 1 round with a muzzle velocity of 905.21 m/s to complete a trajectory of 5.90 m?

### t = d/v (5.90 m / 905.21 m/s) = 0.0065178 s
###
### It would take 0.0065178 s in real life

...

### Calculating how much faster the real-time scene would be...
###
### (2s 167 ms) / (0.0065178 s) = 2.167 / 0.0065178 = 332.47
###
### The special effects mimic a high speed camera
### that captures footage 332.51 times faster than
### real-time. This has the effect that velocities
### appear 332.47 times SLOWER. Therefore, the
### projectile's velocity on film appears to be
###
### 905.21 / 332.47  =  2.72268 m/s

How long does it actually take Superman to stop the bullets by moving himself?

It takes 0 s 458 ms on screen for Superman to move himself in front of the bullets

Distance covered by Superman - 2.83 m

### Distance covered by projectiles in that same time...
### distance = speed*time
###    = 2.72268 * 0.458
###    = 1.246987 m
###
### The ratio of the two distances covered (in the same time)
### tells us how much faster Superman is than the speeding bullet
### (2.83 / 1.246987) = 2.26947 times faster
###
### Therefore, his real-time speed is...
###   2.26947 * 905.21 = 2054.35 m/s

There's an alternative way to calculate this. Superman's speed as the viewer perceives it on the screen is given by

speed = distance/time
= (2.83 m) / (0 s 458 ms)
= 6.17904 m/s

The real-time velocity would be 332.47 times faster, which equates to 6.17904 * 332.47 = 2054.35 m/s

well done, it appears my methodology was way off - thank you

 

[Cubist]

well done, it appears my methodology was way off - thank you

You weren't way off. Making a small mistake anywhere in this type of work can lead to all subsequent results being wrong. I think you've mastered the basics. Experience (practice) helps a lot. You're welcome for the help!