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Bobithan

Member

Joined: 2015-02-15

Posts: 4,476

Smiley Physics

A few questions regarding how a smiley falls...

1. How far does a smiley fall before reaching terminal velocity starting from rest?
2. How long does it take a smiley to reach terminal velocity starting from rest?
3. Does the smiley accelerate uniformly?
4. At terminal velocity, how fast does a smiley fall in blocks/sec?

Exact answers would be appreciated. If the information isn't available I'll try experimenting to get this information on my own, though I'm not sure how to find out if #3 is true or false on my own.

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aka towwl

Luka504

Member

From: Serbia,probs never heard of it

Joined: 2015-02-19

Posts: 2,934

Re: Smiley Physics

I ran some tests in my 400x200 world:my resutls
1.A smiley needs to fall around 120 blocks,after that it keeps falling the same speed till it hits the ground.
2.With no speed and in a straight line,around 50 blocks.
3.Ugh....I dont understand that question,if you meant does it accelarate evenly,then yes,it does.
4.Kinda hard to judge,but around 10-20 blocks per second,thats quite fast

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rgl32

Member

Joined: 2015-02-15

Posts: 543

Re: Smiley Physics

I am doing tests too, and I don't have all same answers as luka504.
So I am counting them like in the image below.

▼Hidden text

So I tried this in a 300x300 world, the most arrows you can break is 15 if you leave one space like in the picture

▼Hidden text

1.) With that, I started with 296 height which does break the arrows, I also got it down to 124, and anywhere between 124 and 150 seems to work completely fine.
     However certain spots above that do not work (some do, some don't), however I have not found a spot where it goes past 15 arrows in a 300x300 world.

2.)A small somewhat inaccurate measurment of time that I did was close to 2 seconds from the 124 mark. Sorry, I couldn't get accurate, a bot will probably help with this. ( I think you meant this, but if you meant horizontal, luka504 was right, but things get weird to farther you go.)

3.) I swear I saw something where someone else tested height velocity and it was not uniform, Ill try to test things and take a pic soon.
3.edit.) I tested some things and it turns out they are pretty uniform

▼Hidden text

also tested like this with the arrows the same height

▼Hidden text

Higher heights might make things turn out weird, but from what I did it seemed uniform. Also, the first 5 arrows should be 4 arrows, I messed up.

4.) Since I got close to 2 seconds, i am going with the smallest number I got so 124/2 = 62 blocks/s. and again, a bot will help with this answer to make it more accurate.

Luka504

Member

From: Serbia,probs never heard of it

Joined: 2015-02-19

Posts: 2,934

Re: Smiley Physics

63 blocks per second seems a bit fast for me
I have done more testing with a friend,i started placing coins over my head and in 1 second we traveled 43 blocks per second
Test 2 has around 46
Test 3 around 51
Its kinda hard to time,making this question difficult to anwser,but somewhere around 40-50 blocks per second
Also its not around 50,around 45 blocks to reach maximum velocity when running.

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Gosha

Member

From: Russia

Joined: 2015-03-15

Posts: 6,211

Re: Smiley Physics

dont forget about glitchs.

rgl32

Member

Joined: 2015-02-15

Posts: 543

Re: Smiley Physics

Also its not around 50,around 45 blocks to reach maximum velocity when running.

I didn't try messing around much with horizontal because there are already numerous people pointing out physics flaws in them (such as not breaking arrows when you should have), and I think the same thing is with vertical, considering you can't break 15 arrows at certain spots above 150.

Luka504

Member

From: Serbia,probs never heard of it

Joined: 2015-02-19

Posts: 2,934

Re: Smiley Physics

dont forget about glitchs.
http://i.imgur.com/OmXAdI6.gif

I seriously doubt that glitch(or any other for that matter) helps at anwsering the questions

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Bobithan

Member

Joined: 2015-02-15

Posts: 4,476

Re: Smiley Physics

So for #3 I recorded myself falling ~100 blocks and looked at the footage frame by frame to see how many coins I collected at certain times and got a graph that looks like this:

Here's the raw data:

▼Hidden text

Time    Blocks Fallen
0.10    1
0.16    2
0.23    4
0.33    6
0.35    8
0.39    9
0.41    11
0.46    12
0.50    14
0.52    16
0.56    18
0.60    20
0.66    24
0.73    29
0.83    35
0.85    36
0.89    40
0.93    41
1.02    46
1.06    51
1.12    54
1.16    57
1.18    60
1.23    62
1.27    65
1.33    72
1.35    73
1.39    77
1.43    78
1.46    80
1.50    83
1.52    86
1.56    90
1.62    93
1.66    98

This line follows the relationship y = 17.828x^2+32.941-5.212 almost perfectly, with y being blocks fallen and x being seconds. I'd suspect the -5.212 comes from my decision for the t = 0 frame being a little off, as it was hard to decide the exact frame you'd consider the smiley to start falling. Remove that and you get y = 17.828x^2+32.941x for the acceleration of a falling smiley.

So for #3 I've gotten the same conclusion. Yes; you do accelerate uniformly.

I probably should've recorded up to ~150 blocks fallen because then we'd probably have a good answer for where exactly you stop accelerating for both time and distance. Maybe I'll do this all again, don't really have anything better to do.

EDIT: Removed the point 0,0 and got better results.

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Different55

Forum Admin

Joined: 2015-02-07

Posts: 16,575

Re: Smiley Physics

y = 17.828x^2+32.941-5.212

Hi could I please know how you came up with this equation.

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Bobithan

Member

Joined: 2015-02-15

Posts: 4,476

Re: Smiley Physics

y = 17.828x^2+32.941-5.212

Hi could I please know how you came up with this equation.

Google Spreadsheets/Excel can give you a line of best fit. That equation follows the red trendline.

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aka towwl

imgood9

Member

From: 'Murica

Joined: 2015-02-28

Posts: 472

Re: Smiley Physics

I looked at the footage frame by frame to see how many coins I collected at certain times and got a graph that looks like this:
http://i.imgur.com/Lls78Ma.png

Here's the raw data:

▼Hidden text

Time    Blocks Fallen
0.10    1
0.16    2
0.23    4
0.33    6
0.35    8
0.39    9
0.41    11
0.46    12
0.50    14
0.52    16
0.56    18
0.60    20
0.66    24
0.73    29
0.83    35
0.85    36
0.89    40
0.93    41
1.02    46
1.06    51
1.12    54
1.16    57
1.18    60
1.23    62
1.27    65
1.33    72
1.35    73
1.39    77
1.43    78
1.46    80
1.50    83
1.52    86
1.56    90
1.62    93
1.66    98

This line follows the relationship y = 17.828x^2+32.941-5.212 almost perfectly, with y being blocks fallen and x being seconds. I'd suspect the -5.212 comes from my decision for the t = 0 frame being a little off, as it was hard to decide the exact frame you'd consider the smiley to start falling. Remove that and you get y = 17.828x^2+32.941x for the acceleration of a falling smiley.

So for #3 I've gotten the same conclusion. Yes; you do accelerate uniformly.

Expanding on your work, considering this is the position with respect to time you can find your velocity and acceleration functions as well.

▼Deriving Velocity/Acceleration Functions from Bobithan's Position Function

y = 17.828x^2 + 32.941x relates position (p) with respect to time (t).
it follows that dp/dt = v(t) = 2(17.828)x^(2-1) + (1)32.941x^(1-1)
that simplifies to our velocity equation: v(t) = 35.656x + 32.941.

To find the acceleration function, we repeat.
dv/dt = a(t) = 35.656

v(t) = 35.656x + 32.941 models the rate at which your smiley is changing position for "x" seconds. (result in blocks/second)
a(t) = 35.656 being the rate of acceleration, up to a certain point. This is assuming you never reach terminal velocity.

Assuming Luka504's estimate of 45 blocks/second is true, you'll accelerate for a certain amount of time before you stop.
So when do you stop accelerating?

▼Algebra

v(t) = 35.656x + 32.941
45 = 35.656x + 32.941
12.059 = 35.656x

x = 0.338 seconds.

Seems low. How about we find x for v(t) = 60?
60 = 35.656x + 32.941
27.059 = 35.656x

x = 0.758 seconds.

x = 0.338 seconds. if the terminal velocity is 45 blocks/second.

Okay, so we have a lower bound. What if the terminal velocity is 60 blocks/second?

x = 0.758 seconds. if the terminal velocity is 60 blocks/second.

These are the theoretical lower/upper bounds for the time taken to reach terminal velocity in EE. In other words, these are the times it takes to stop accelerating, assuming linear acceleration of 35.656 blocks/second/second.

My main question, however, is how the speed increase/low gravity/high jump power-ups impact the game's physics and alter these equations.

Tomahawk

Forum Mod

From: UK

Joined: 2015-02-18

Posts: 2,847

Re: Smiley Physics

Decompile the swf. You can probably calculate the answers from the data in there.

Alternatively, fall while spamming keys, and record all the movement packets with a bot, which will provide things like distance travelled and instantaneous velocities. Work from there.

You can roughly answer Qs 1&2 by placing the number of up arrows required to stop a player falling at full speed, and dropping onto them from various heights.

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Minisaurus

Banned

Joined: 2021-04-26

Posts: 62

Re: Smiley Physics

I just tested 4# and got a result of around 84 blocks per second in terminal speed fall
Now that EE is "finished" product, is the fall speed 84 blocks per second in terminal speed?

I also tested horizontal speed and got a result of around 42 blocks per second (half the speed of fall, friction seems like reducing the speed in half)

Used cursed action tool and let 4 seconds for acceleration and one second to take coins, the amount of coins taken in the last second is the number I used to calculate blocks per second at terminal speed.

Kirby

Member

Joined: 2015-04-04

Posts: 4,307

Re: Smiley Physics

A few questions regarding how a smiley falls...

1. How far does a smiley fall before reaching terminal velocity starting from rest?
2. How long does it take a smiley to reach terminal velocity starting from rest?
3. Does the smiley accelerate uniformly?
4. At terminal velocity, how fast does a smiley fall in blocks/sec?

Exact answers would be appreciated. If the information isn't available I'll try experimenting to get this information on my own, though I'm not sure how to find out if #3 is true or false on my own.

Ik this post is 6 years old but

1) About 1447 blocks
2) About 727 Blocks
3) Yep
4) Terminal fall velocity is 13.553105760940054 pixels per tick, or 84.7069110059 blocks per second
    Terminal run velocity is 6.776552880470027 pixels per tick, or 42.3534555029 blocks per second (terminal run velocity = terminal fall velocity/2)

NoNK

Member

Joined: 2019-07-13

Posts: 922

Re: Smiley Physics

nreat