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Look At This Calculation
Started by: SquallX

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SquallX
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Location: miami/fl.

Look At This Calculation

http://www.narutoforums.com/showthread.php?t=715365

Could someone check out this calculation. It was made to calculate Gai's speed, but it was locked not because the calculation was wrong, but because no one in the OBD could debunked it. So instead they just troll, and made a mod locked it.

So could anyone check it, and see if it's wrong or right. Or is the OBD just bias.

Old Post Apr 6th, 2011 06:48 AM
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Q99
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One issue is we don't know exactly when Gai started moving. So it's probably on the high-end side.


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Old Post Apr 6th, 2011 07:09 AM
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dadudemon
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Another issue I have is any sort of time frame reference. They keep referring to the shuriken but I'm not seeing that as usable, at all.

We have no idea how fast Gai is because we have no time unit in reference to the events in that frame sequence.

It could have been highly stylized in ultra slow-mo (actually, that's exactly what was supposed to be happening) and then a time skip to show that Gai and the proctor stopped the rest...which is actually what happened.


That's the only issue I have with that: we really need a tangible time reference in order to make a comparison. IMO, his calculation is completely baseless.



The only time reference we have in Naruto that gives us a nice speed is the 5 second cool down sequence at the very end of Naruto vs. Pain. That was an excellent way to figure out velocity.


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Old Post Apr 6th, 2011 07:34 AM
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BloodRain
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[Using the same method as him]
Found a distance of 12.44m traveled, imo the red lines should of been measured at the base line by Kakashi's feet. √(5.52m^2+11.15m^2)

Have yet to see a source saying a kick is 50m/s, 'til then I'm going with 30m/s that google says. Others in that thread say between 25m/s-35m/s.

0.19/30 = 0.00633 sec
12.44/0.00633 = 1965.24487 m/s

Thats Mach 5.77.

Main issue I have with this is http://www.mangareader.net/93-91-9/...chapter-86.html the bottom panal that has the sand much further away from Lee then in that scene. The perspective may be different in the used image. That and if the sand was at full speed in that grasp seeing as Gaara was beat down and had no reason to use a fast attack on a half-paralyzed Lee. The anime would support this^ image and also that the sand was not at its fastest.

If it is a perspective issue and the bottom panel is in fact correct [around 1m instead of 19cm] then the speed only ends up as Mach 1.


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Old Post Apr 6th, 2011 12:34 PM
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dadudemon
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quote: (post)
Originally posted by BloodRain
[Using the same method as him]
Found a distance of 12.44m traveled, imo the red lines should of been measured at the base line by Kakashi's feet. √(5.52m^2+11.15m^2)

Have yet to see a source saying a kick is 50m/s, 'til then I'm going with 30m/s that google says. Others in that thread say between 25m/s-35m/s.

0.19/30 = 0.00633 sec
12.44/0.00633 = 1965.24487 m/s

Thats Mach 5.77.

Main issue I have with this is http://www.mangareader.net/93-91-9/...chapter-86.html the bottom panal that has the sand much further away from Lee then in that scene. The perspective may be different in the used image. That and if the sand was at full speed in that grasp seeing as Gaara was beat down and had no reason to use a fast attack on a half-paralyzed Lee. The anime would support this^ image and also that the sand was not at its fastest.

If it is a perspective issue and the bottom panel is in fact correct [around 1m instead of 19cm] then the speed only ends up as Mach 1.


There's also the fact that Lee's arm and leg are not resting on the ground which means gravity would have had to take over and pull it to the ground. Since he's flying across the air, his arm and leg are floating, allowing us to use the gravitational acceleration constant to see how long it would take for them to "fall" to the earth after he stops flying horizontally. We can ignore that his leg and arm are actually falling about a fulcrum because his arm and leg, at the very ends, would fall at the gravitational constant and this is where we are going to make our measure from. Trust me, it works out. smile


We can do a very easy calculation for how long it would take for the highest point to fall all the way to the ground because gravity is a constant (which we will assume he is at sea-level, lol)

So, doing the math, we will finally have an exact answer which you can post everywhere on all of the internets:


First, we need to figure out the dimensions. We know Lee's height because we have that information.

So, we can do a pixels to heigh comparison.

We have to use a bit of highschool geometry, though, to get the height of the foot and compare it to how tall Lee is in order to come to a conclusion of a pixel per unit measure.


So, here's the image, measured with the parts broken down (we will be using the Pythagorean theorem to get the hypotenuse measures for the two triangles.)

(please log in to view the image)

First, we'll do the long one:

(please log in to view the image)

It has a height of 188 pixels and a width of 582 pixels.

The hypotenuse is, therefore:

(188^2) + (582^2) = x^2

35344 + 338724 = x^2

374068 = x^2

374068^(1/2) = x

x = 611.6 ~ 611

So that measure is 611 pixels. I rounded down due to the angles making Lee taller than he actually would be.




Next up, we will do the short one:

(please log in to view the image)

Height is 80 pixels and width is 104 pixels.

(80^2) + (104^2) = x^2
6400 + 10816 = x^2
17216 = x^2
17216^(1/2) = x
x = 131.2 or ~ 131 pixels

Add them together to get his height in pixels:

131+611 = 742 pixels


His height is 158.5 cm.

So, for every pixel, he is 0.21 cm/pixel. Again, I rounded down due to the angles I used to measure his height.


So now we need to find the height from the top of his foot to the ground:

(please log in to view the image)


It is a height of 265 pixels.

Multiply your pixels by your cm conversion factor to get those pixels in cm:

265pixels*.21cm/pixel = 55.65cm




We are VERY close, now, to figuring out a time.

Now we need to figure out how long it takes, under the gravity constant, to fall 55.65cm.


d=(v_0*t) + (1/2)*(a)*(t*^2)

d = 55.65cm

v_0 = 0

a = 9.806 m/s/s


No substitute your variables:

55.65cm = (v_0*0) + (1/2)*(9.806m/s/s)*(t^2)

Simplify

55.65cm = (1/2)*(9.806m/s/s)*(t^2)

Simplify again:

55.65cm = (4.903m/s/s)*(t^2)

Almost done simplifying:

55.65cm/(4.903m/s/s) = t^2 which is also .5565m/4.903m/s/s=t^2

Final simplification:

0.1135019375892310830104017948195s/s = t^2

Square root both sides to get your time:

0.337 seconds.





So we now have a time to do the calculation by.



This time is a bare minimum time.


We can see here that Gai moves sometime between the panel I used to calculate height and the last panel where Lee's feet are resting and we can see that the sand was dispersed...so Gai would be faster than the number we come up with, but it should be close enough:

People are saying he traveled 15 meters in .336 seconds.

That's 44m/s.






But some may cry that Gai moves AFTER Lee's feet start to fall: divide the time by half, if you want to, but that's rather arbitrary. We don't know if those panels inbetween were concurrent with Lee hitting the ground, if it occured after Lee hit the ground, or not.


So we will just pretend it's halfway inbetween, my final answer is:

88m/s is his average speed to cross that distance.



What is 88m/s in terms of the speed of sound at sea-level?

340.29 m/s = SoS

25.8% the speed of sound for Gai's speed feat.







Post this post everyone an every Naruto forum, everywhere. no expression


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Last edited by dadudemon on Apr 7th, 2011 at 05:31 AM

Old Post Apr 7th, 2011 05:23 AM
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