User blog:ZeroTwo64/User:ZeroTwo64/Game & Watch Zelda calculations

In this blog, I'm going to explain some calculations I did about Game&Watch Zelda.

According to Saikou The Lewd King's calculations, the platform in which Link fights his enemies is 537 pixels or 130 mm.

According to Online Pixel DPI Calculator the Lenght of the Room is 992 pixels or 240 mm, while its height is 620 pixels or 150 mm. So the room is 992x620 pixels.

Link's lenght is 62 pixels or 15 mm, and his height is 124 pixels or 30 mm.

I considered Young Link's height in Ocarina of Time, that is around 5'1 or 1.30 m, so his lenght is 0.65 m.

Now I found the size of the room in proportion of Link's:

ROOM'S HEIGHT=  1.3 m : 124 pixels = y: 620 pixels;  y =(620 x 1.3) / 124= 6.5 m

ROOM'S LENGHT=  0.65m : 62 pixels = y: 992 pixels; y=(992 x 0.65) / 62= 10.4 m



In the second picture I used, the main room's lenght is 105 mm, while the lenght of the dragon's room is 60 mm, so it should be 137 mm in the first picture.

DRAGON'S ROOM LENGHT (in mm)=  105 mm : 240 mm = 60 mm : y;  y= (240 x 60) /105= 137 mm

DRAGON'S ROOM LENGHT (in m)= 240 mm: 10.4 m = 137 mm : y;    y= (137 x 10.4) /240= 5.94 m

Dragon's room height is equal to main room's so it is 6.5 m.

The main room in the second picture is 434 x 289 pixels

The Dragon's height is 165 pixels or 40 mm.

DRAGON'S EIGHT (in m)=  165 pixels : 289 pixels = y : 6.5 m;   y= (165 x 6.5) /289= 3.7 m

The last dungeon's lenght is that of 4 rooms, while its height is 5 rooms+Dragon's room, whose height is equal to normal room's.

DUNGEON'S LENGHT= (4 x 10.4 m) = 41.6 m

DUNGEON'S HEIGHT= (6 x 6.5m) = 39 m

DUNGEON'S SIZE= (41.6 x 39) m

At this point I calculated the distance travelled by the arrow in meters: 992 pixels : 10.4 m = 537 pixels : y;  y= (10.4 x 537) /992= 5.63 m

The speed of the spear is 41.67 m/s, so it needs 0.135 s to travel 5.63 m;  time= space/speed=5.63/41.67=0.135 s

Now, I used 2 others picture to calculate the speed of the spear in another prospective:



The distance between Link and the Moblin is 10 mm or 4.3 m; 10mm:y=13 mm:5.63 m;

y= (10 x 5.63)/13= 4.3 m

For the spear it tooks 1.62 s to go from the moblin to Link, so its speed is 2.65 m/s;

speed=4.3m/1.62s= 2.65 m/s

I used 2 pictures to calculate the speed of Dragon's fireball:



It tooks 3.14 s for the fireball to reach the end of the screen and it travelled for 5.5 m; in order to travel for 4.3 m it needs 2.43 s;     5.55 m : 3.14 s = 4.3 m : y;

y=(3.14 x 4.3)/ 5.55= 2.43 s

Its speed in this system is 1.77 m/s;   speed=4.3 m/2.43 s= 1.77 m/s

Its speed in the first system I used is 27.8 m/s;  2.65m/s : 41.67m/s = 1.77 m/s : y;

DRAGON'S FIREBALL SPEED= (41.67 x 1.77)/2.65= 27.8 m/s

According to Wikipedia, the mass of the rock used in ancient buildings varies from 2300 Kg/m^3 to 2700 Kg/m^3. I used the middle value= 2500 Kg/m^3

Now I calculate the energy the Dragon need to destroy its dungeon:

KE=0.5 x Mass x (Speed^2)=0.5 Volume x Density x (Speed^2)=

0,5 x Density x Height x 3,14 x (Radius^2) x (Speed^2)

0.5 x 2500 x 39 x 3.14. (20.8^2) x (27.8^2)=5.46x10^10 J or 13.05 tons

8-B City Block Level