Frame Rate Questions

[koopatroopa]

Hey

So this is sorta a basic question:
Until now I have been doing all my coding in the love.update() function but i began to think "If this was being played on a much faster computer that could manage a faster frame rate wouldn't the game be played much quicker to because it would be updating far faster" ?

if i say as long as the mouse is down change X by 3 this will happen every frame but if the frame rate is higher it will happen faster???

if this is True how do you work around it

[ivan]

Yes, you have to take into account delta(dt) which is the time elapsed between updates.
So instead of change X by 3 you would write change X by (3*dt)
where 3 is the speed (3 units, usually pixels, per second)
and dt is delta in seconds

[Ellohir]

dt is the elapsed time. So, if one call is made 0.1 seconds after the previous, that code would move the player 0.3 pixels on position; so, in the end, after all the different times we can have, the player will always move 3 pixels on one second, whatever machine it is played on

[verilog]

Hey folks,

I've a similar question. Suppose you have a function which contains an array and a counter to drive its index. The function is called randomly by the user. The functionality you’re looking for is to cycle trough the array elements each time a frame is drawn, that is, we get element number 1 (index = 1) of the array at frame 1, we get element number 2 (index = 2) at frame 2, and so on.

Here, we need the counter to update after every dt time, but the counter can’t be a direct function of dt, as we strictly need integer values (incremented in exactly 1 unit) to drive the indices of our array. How would you do that?

[MarekkPie]

If I understand correctly, then can't you just increment the counter in love.update(dt)?

Code: Select all

array = {--[[stuff]]}

function A()
  return array[index]
end

function love.update(dt)
  index = index + 1
end

I'm a little confused by what exactly you're looking for. Give an example.

[verilog]

Hey MarekkPie,

I’ll tell you why I’m asking this. In the game I’m working on, I wanted to give the illusion of a “smooth” character jump; I decided to move the player along the Y axis according to a free fall formula, a parabola.

Anyway, I computed the Y points needed for this and stored them in an array, so each time the “jump” function was called the character was translated to a Y position according to the Y values in my array. The character needed to move one position exactly on each new frame, so I was a little bit unsure of how to do this, hence my previous question.

What I ended up doing was exactly what you posted, but I was wondering if there is an alternate method for achieving this.

[Robin]

You could use dt.

Code: Select all

function A()
  return array[math.floor(index)] -- this is the trick that makes it work
  -- it rounds index to the nearest smaller-or-equal integer.
end

function love.update(dt)
  index = index + dt * speed
end

See math.floor for more information.

[miko]

Here, we need the counter to update after every dt time, but the counter can’t be a direct function of dt, as we strictly need integer values (incremented in exactly 1 unit) to drive the indices of our array. How would you do that?

The counter can be float, only array index should be integer. So:

Code: Select all

char=Char[math.floor(counter)]

would do it.

[MarekkPie]

Depending on his system speed, just flooring dt might skip or repeat frames and whatnot. For example, an average dt of 1.5 would give him:

Code: Select all

First pass:
array[math.floor(dt)] or array[1]

Second pass:
array[math.floor(dt * 2)] or array[3]

I think the only way to guarantee a number will give you the correct integer value is if you increment it by an integer. At which point just having a separate integer-based counter is best.

[Robin]

Depending on his system speed, just flooring dt might skip or repeat frames and whatnot. For example, an average dt of 1.5 would give him:

You know, if dt is 1.5, you have an FPS of 0.67. Skipping frames is not your worst problem then.