Originally posted by BloodRain
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/naruto/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 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.
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.)
http://i191.photobucket.com/albums/z238/dadudemon/LeeMaster.jpg
First, we'll do the long one:
http://i191.photobucket.com/albums/z238/dadudemon/LeeLong.jpg
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:
http://i191.photobucket.com/albums/z238/dadudemon/LeeShort.jpg
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:
http://i191.photobucket.com/albums/z238/dadudemon/LeeHeight.jpg
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.
