Dragon Ball Manga - Tao Pai Pai Jumps to His Pillar

Thanks to @NemesisReloaded. Dragob Ball Chapter 85 See New Calculation 3

Calculation 1 (Old)[]

Pillar Speed[]

After being brought in to deal with Goku and retrieve the Dragonballs in his possession [DB #85], Mercenary Tao breaks off a stone/marble/granite/concrete pillar and throws it 2,300km to Goku's location.

Media:original-11130-111307308-5642717-merctaothrowa.jpg

Media:original-11130-111307308-5642757-merctaothrowc.png

When he does this, Tao is standing on an upper terrace of the Red Ribbon Army Head Quarters. The following is a grouping of more of less all of the pictures of the HQ complex from the outside. In some you can see the pillar that Mercenary Tao removed.

Media:original-11130-111307308-5642727-merctaothrow2.png

Thankfully, people much smarter than me at convertalot.com created a ballistic trajectory calculator using a lot of equations, that all work out visually something like this:

Media:original-11130-111307308-5642739-merctaothrow3a.png

If you know initial speed of the projectile, the height the projectile was fired at and the angle it was fired at, it'll tell you the range. Well, we know the range, we want the speed. However, the speed can be guessed at with a game of higher or lower IF we know the height the projectile was fired from, and the angle.

Height:[]

Media:original-11130-111307308-5642827-merctaothrow1.png

Goku is in Karin's Sacred Land [DB #84], 2,300km from the Red Ribbon Army Headquarters. Both are flat lands situated near mountains. They are possibly similar heights from sea level. It's all I have to go on, so I will assume the only difference between them is the height of the building Tao is in. Not that I imagine it would make a huge difference, but I may as well pin down as many variables as I can.

Looking at the above, there is little doubt that the follow is the front of the complex, located at the main gate:

Media:scale-super-11130-111307308-5642730-merctaothrow2a.png

But what of the terrace that Tao threw the pillar from? There are SOME clues. The large domed building at 'A' and the building with the cannon on it at 'B'

Media:original-11130-111307308-5642761-merctaothrow2b.png

Media:original-11130-111307308-5642764-merctaothrow2d.png

And a view of the main complex from the other side:

Media:original-11130-111307308-5642765-merctaothrow2c.png

Which means that this building

Media:original-11130-111307308-5642770-merctaothrow2e.png

Is this building:

Media:original-11130-111307308-5642771-merctaothrow2f.png

Now we know roughly how far from the ground we are. The roof of the building on the lower right is about 2 floors from the ground. Luckily because Akira Toriyama uses perspective lines in this picture, we can figure out roughly where the ground is and knowing the size of the column we can figure out how high he was when he threw it.

Media:original-11130-111307308-5642831-merctaothrow2g.png

Given that there'll be a slight variation on that because of the drawing and the perspective, a nice round 50m sounds ok by me.

Angle:[]

One of the pictures in DB #85 is virtually side view, and is immediately after Toa throws and lands on the pillar, making gaining an angle fairly simple.

Media:original-11130-111307308-5642832-merctaothrow3b.png

Final Speed:[]

Using the calculator at http://www.convertalot.com/ballistic_trajectory_calculator.html the initial speed of the pillar must be 7049.453 m/s. This means that Mercenary Tao's hand must have traveled 7049.453 to throw it. The equivalent of 25,378 km/h or 15,769 mph. Just over Mach 20. But it's worth remembering that he's throwing a stone/concrete pillar, which is not light. Unladen, his hand should move faster, so...

Mercenary Tao can move his limbs in excess of 15,769mph

Pillar AP[]

While this is a throwing or movement feat, Tao is throwing an object that has a mass. If we can find the mass of the pillar he threw and the acceleration it underwent, we can figure out the force that was applied to it by him.

Mass:[]

Media:original-11130-111307308-5643068-merctaothrow1.png

The pillar itself has a volume of v = h x pi x r^2 which is 0.3026 cubic metres. But what's it made of?

Well, it's not wood or steel. The way the ends are damaged suggest some kind of stone, plaster or concrete. So looking at the densities of a range of materials, I found the following densities:

Material	Density (1000kg/m^3)
Marble	2.70
Granite	2.70
Limestone	2.56
Concrete	2.40
Sandstone	2.20
Plaster	0.85

This means that the pillar could weigh as much as 817kg or as little as 257kg. Given the nature of the building as a military purpose build structure, as well as having prison facities and a need to be strong, I personally find that concrete, being cheaper and harder, is the most likely used material, which would put it at 726kg.

Acceleration:[]

Acceleration can be determined by final velocity minus initial velocity, divided by the time it took to accelerate, or a = (v-u)/t

The Initial Velocity is 0m/s as it's at a stand still
The Final Velocity is 7049.453m/s at the point it left Tao's hand
But how much time did it take to get there?

We know the initial speed is 0m/s when he started the throw, and the final speed is 7049.453m/s when he released it. That means over a given distance, and an assumed constant acceleration, it had an average speed of 3524.7265m/s for the duration of the throw. If we find the distance Tao's hand moved during the throw, we'll know how long the throw lasted for.

Knowing General Tao's height to be 1.78m [Daizenshuu 7] some of the pictures can be adapted to allow for a reasonable idea of the length of his throw. The picture of his throwing stance needs to be skewed to present something closer to a side view, instead of the 3/4 rear view and another full-body picture used to fill out the rest of his body.

Media:original-11130-111307308-5643151-merctaothrow3c.png

Using these pictures, the line of the throw and the natural joints of the body, it's possible to gain an idea of Tao's movements in the throw.

Media:original-11130-111307308-5643158-merctaothrow3d.png

Given the potential for inaccuracy in the picture, I will round to 1.6m

Since Speed = Distance / Time, then Time = Distance / Speed. Therefore the duration of this throw was 0.000454s.

Therefore the Acceleration of the pillar is = 15,527,430m/s/s

Force:[]

Force = Mass x Acceleration

The range of values for the linear force involved in accelerating this pillar is:

Pillar Material	Mass (kg)	Force (N)
Granite (Highball)	817	12,685,909,914
Concrete (Likely)	726	11,272,913,828
Plaster (Lowball)	257	3,990,549,385

Mercenary Tao's strike force is between 4 and 12.7 gigaNewtons, likely to be 11.3 gigaNewtons.

Calculation 2 (Tao Pai Pai Speed)[]

In DB #85 when Mercenary Tao appears to leap from the terrace to the pillar he had just thrown he presents an opportunity to calculate a speed feat. Mercenary Tao is the brother of Master Shen of the Crane School of Martial Arts, later to include Tien and Chiao-Tsu. A common technique learned by the students of this school along with the Dodon Ray (first used in Dragonball by Mercenary Tao against Goku [DB #86]) is the Bukujutsu, or Sky Dance - otherwise known as flying.

It's unknown if Mercenary Tao can actually fly because he never does, but he might be able to and is never specifically shown to fly. So it's impossible to know whether or not this the act of landing on the thrown pillar is simply a leap, or if it's more akin to the speed burst many fighters use at the beginning of about to reduce the distance quickly to make range attacks more difficult. If it IS a leap does that count as a speed of limb movement? For me, I feel this is too similar to a speed burst at the start of about, not just a leap or flight, the stance is all wrong, but it's open to interpretation.

If this is a leap, you have to wonder why he doesn't just jump the 2,300km, especially since it's within his capability as he could leap fast enough to catch up with and land on the pillar he threw. I think this is to do with the landing. Before the pillar lands near Goku Tao jumps up in the air. It seems that the reason he may not fly might be energy conservation, and the reason he rides the pillar instead of just jumping to Goku's location is to not end up buried in the ground like the pillar was.

In DB #85 Tao throws the pillar and virtually immediately sets off to land on it.

Media:original-11130-111307308-5645309-merctaospeeda.jpg

The first and third frames are drawn using perspective lines, though they aren't perfect, so I compensated for where the lines (particularly on the 3rd frame) should have gone. We know the dimensions of the pillar, and because of the perspective, I can use those dimensions to find the distance it has traveled.

Media:original-11130-111307308-5645330-merctaothrow1.png

Media:original-11130-111307308-5645331-merctaospeed1a.png

You can find the perspective distance by calculating intervals of known distances. I used the difference in the diameter of the pillar from both ends to find a ratio. I used this ratio to find the apparent dimensions of the pillar at one stage closer, and so on. In this picture, the pillar has traveled a distance of 4.53m. Since the pillar is moving at a speed of 7049.453 m/s, this picture exists at an instant 0.00064s after Tao released it.

Media:scale-super-11130-111307308-5645359-merctaospeed1.jpg

In the above picture, there are measurably 17 segments that equate to 2.408m long between the back of the pillar and where Tao released it. I corrected the perspective of this picture slightly because the lines lead to the wrong point, though close. So I adjusted the lines to where Tao was to continue the calculation. At 17 segments, this means that the pillar is 40.936m from where Tao released it or 0.00581s after Tao released it. Assuming Tao moved immediately after Frame 1 (throwing the pillar) Tao has been in the air for at most 0.00516s. This is too low. Tao is 1.49m behind where he should be.

Tao himself is between the 14th and 15th pillar segments, or 34.92m from where the pillar was released. This pillar took just 0.004953s to get to that same point. For Tao to catch up with the pillar, he must be faster than the pillar and have covered the same distance in less time, not more. This means he did not leave immediately after Frame 1. I need to account for the length of Tao's arm.

Media:original-11130-111307308-5645390-merctaospeed2.png

Tao's arm length means he has had to move an extra 0.74m to catch up to the pillar. This puts his speed at 6910.853 m/s which is still too slow to catch up with the pillar, which he clearly did. The difference between how far behind the pillar Tao was in Frame 1 and Frame 3, including arm length, is 0.75m which means Tao paused for 0.000106s before he launched, and that's for equal speed with the pillar. Perhaps it's impossible to discover how fast Tao did jump, but we can put a maximum on it.

If we assume that Mercenary Tao landed on the pillar on its next segment of travel, as is possible given the manga:

Media:original-11130-111307308-5645405-merctaothrowb.jpg

This would mean that Tao would have covered the distance of 4 segments or 9.632m in 0.000342s, which is 28,197.812 m/s. This is the absolute maximum speed Tao could be traveling, and likely to be too fast since it's the extreme scenario. At this speed, Tao would have taken 0.001264s to travel the distance of 35.656m that Frame 3 shows, meaning he would have paused 0.00455s before launching himself at the pillar.

It makes sense that since Tao's feet start below where the pillar started off that for him to land on the pillar his own trajectory would be a tighter arc. It's worth remembering that in manga or anime, things look how the artist best sees fit to display the image. That's why pictures use perspective lines. With that in mind, artistically, this is where I feel that Tao would land as compared with this picture:

Media:scale-super-11130-111307308-5645414-merctaospeed3.jpg

Which, if correct (and why wouldn't it be? lol) would mean that Tao would land on the pillar after another 3 segments traveled, or another 0.001025s:

Media:scale-super-11130-111307308-5645754-merctaospeed3a.jpg

Tao would have traveled 6 pillar lengths (14.448m) in 0.001025s, giving him a burst speed of 14,098.906m/s, which is twice the speed of the pillar, which was a happy accident considering the arc was drawn before the segments were calculated. But I'd still be more comfortable with a Lowball figure too.

Frame 3 is when the pillar was 40.936m or 0.005807s after Tao threw it. IF the pillar had traveled a not unreasonable total of 200m when Tao landed on it (5x farther from the terrace than Frame 3), from the time of the throw the pillar will have been traveling for 0.028371s. Therefore there is a difference of 0.022564s between Tao's position in Frame 3 (7.224m behind his landing position on the pillar) and landing on the pillar at 200m. The pillar itself has another 159.064m to travel before Tao lands on it, or near exactly 66 segments. Meaning Tao has to travel 69 segment lengths (from Frame 3) in 0.022564s to land on the pillar. Which means he would have had to have traveled at a speed of 7363.581m/s to have caught up with it.

Mercenary Tao's Burst Speed is:

Lowball - 7,363.581m/s or Mach 21.46 Hypersonic+
Artistic - 14,098.906m/s or Mach 41.10 High Hypersonic
Highball (max) - 28,197.812m/s or Mach 82.20 High Hypersonic+

Calculation 3 (New)[]

Initial Values[]

Pillar Travel Speed = 7070.45975424530957 m/s

https://www.amesweb.info/Physics/Projectile-Motion-Calculator.aspx

In DB #85 Tao throws the pillar and virtually immediately sets off to land on it.

Media:original-11130-111307308-5645309-merctaospeeda.jpg

The first and third frames are drawn using perspective lines, though they aren't perfect, so I compensated for where the lines (particularly on the 3rd frame) should have gone. We know the dimensions of the pillar, and because of the perspective, I can use those dimensions to find the distance it has traveled.

Media:original-11130-111307308-5645330-merctaothrow1.png

Media:original-11130-111307308-5645331-merctaospeed1a.png

You can find the perspective distance by calculating intervals of known distances. I used the difference in the diameter of the pillar from both ends to find a ratio. I used this ratio to find the apparent dimensions of the pillar at one stage closer, and so on.

In this picture:

Pillar has traveled a distance of 4.53 m.

Pillar is speed = 7070.45975424530957 m/s,

Time after release = 0.000640693838513 s

Media:original-11130-111307308-5645359-merctaospeed1.jpg

In the above picture, there are measurably 17 segments that equate to 2.408m long between the back of the pillar and where Tao released it. I corrected the perspective of this picture slightly because the lines lead to the wrong point, though close. So I adjusted the lines to where Tao was to continue the calculation.

At 17 segments:

Pillar distance from release = 40.936 m

Pillar is speed = 7070.45975424530957 m/s

Time since release = 0.00578972251068 s

0.00578972251068 - 0.000640693838513 = 0.00514902867217 s

Assuming Tao moved immediately after throwing the pillar Tao has been in the air for at most 0.00514902867217 s.

This time is too high.

Tao himself is at the 14th pillar segments

Tao distance from release = 33.712 m

This pillar took just 0.0047680067735 s to get to that same point.

For Tao to catch up with the pillar, he must be faster than the pillar and have covered the same distance in less time, not more.

This means he did not leave immediately after Frame 1.

Let's account for the length of Tao's arm.

Media:original-11130-111307308-5645390-merctaospeed2.png

[18]

Tao's arm length means he has had to move an extra 0.74 m to catch up to the pillar.

(34.92+.74)/0.00514902867217 = 6925.57805955 m/s

This puts his speed at 6925.57805955 m/s This is still too slow to catch up with the pillar, which he clearly did.

Frame 1 Gap = 4.53 m

Frame 3 Gap = 40.936 - (34.92 + .74) = 6.484 m

Gap Difference = 1.954 m

The difference between how far behind the pillar Tao was in Frame 1 and Frame 3, including arm length, is 0.746 m.

1.954 / 7070.45975424530957 = 0.000276361095023

This means Tao paused for 0.000105509404753 s before he jumped, and that's for equal speed with the pillar.

Maximum[]

Perhaps it's impossible to discover how fast Tao did jump, but we can put a maximum on it.

If we assume that Mercenary Tao landed on the pillar on its next segment of travel, as is possible given the manga:

Media:original-11130-111307308-5645405-merctaothrowb.jpg

[19]

Rear End of Pillar distance from release = 40.936 m

Tao Pai distacne from release = 33.712 m

Pillar Moves Additional = 1 Length = 2.408 m

Additional Time = 0.000340571912393 s

Gap with Tao = 9.632 m

Tao's Speed to reach piller = 28281.839017 m/s.

This is the absolute maximum speed Tao could be traveling, and likely to be too fast since it's the extreme scenario. At this speed, Tao would have taken 0.00119200169337 seconds to travel the distance of 33.712 meters that Frame 3 shows, meaning he would have paused 0.00459772081731 seconds before launching himself at the pillar.

It makes sense that since Tao's feet start below where the pillar started off that for him to land on the pillar his own trajectory would be a tighter arc.

Artistic[]

It's worth remembering that in manga or anime, things look how the artist best sees fit to display the image. That's why pictures use perspective lines. With that in mind, artistically, this is where I feel that Tao would land as compared with this picture:

[20]

Which, if correct (and why wouldn't it be? lol) would mean that Tao would land on the pillar after another 3 segments traveled, or another 0.00102171573718 seconds:

[21]

Tao would have traveled 6 pillar lengths (14.448m) in 0.00102171573718 seconds

giving him a burst speed of 14140.9195085 m/s

which is twice the speed of the pillar, which was a happy accident considering the arc was drawn before the segments were calculated.

Lowball

But I'd still be more comfortable with a Lowball figure too.

Frame 3 is when the pillar was 40.936m or 0.00578972251068 seconds after Tao threw it.

IF the pillar had traveled a not unreasonable total of 200m when Tao landed on it (5x farther from the terrace than Frame 3), from the time of the throw the pillar will have been traveling for 0.0282867036871 seconds.

Therefore there is a difference of 0.022564s between Tao's position in Frame 3 (7.224 meters behind his landing position on the pillar) and landing on the pillar at 200m. The pillar itself has another 159.064 m to travel before Tao lands on it, or near exactly 66 segments. Meaning Tao has to travel 69 segment lengths (from Frame 3) in 0.022564s to land on the pillar. Which means he would have had to have traveled at a speed of 7363.581m/s to have caught up with it.