Need some help with dt

Hello,

I'm new to the scirra forums and construct2 but recently joined after someone showed me some cool things you can make with the game engine. But to get to the point, I have some questions about dt.

I think I have a basic understanding of how its used but I can't seem to fully grasp it. Lets say I want to move an object like this:

Event: Every Tick--> Sprite set Y--> to sprite.Y+5

I know there are behaviors that come with dt, but I would like to move the object this way for a scenario in the game. I would like to move the object 5 pixels down. The problem is it's not always fast unless I add dt but even so I can't seem to get it right. I would like for it to run as fast as possible on different systems and devices but have it move at 5 pixels.

So I was wondering if I would set the sprite's Y position to sprite.Y + (300*dt)

Or would it be sprite.Y + (500*dt)?

I assumed it would be 300 since multiplying 5*60 gives 300. So every 60 seconds the sprite would move 300 pixels which would be 5 pixels a second?

Or am I wrong? I find dt a bit confusing so I would appreciate it if someone could tell me how I can move a sprite object by 5 with dt. (Sprite.Y+5)

If in 60 seconds you wanted to move 300 pixels the equation you want is sprite.Y + (300/60*dt) = sprite.Y + (5*dt)

dt is the time that has passed since the last frame so in 1 second the sum of dt will equal 1

Does that make sense?

LevelUpJordan

Hello, I appreciate your example it does make alot more sense (I think). So does that mean if I change the equation to sprite.Y + (300/30*dt), then the sprite object will move 300 pixels in 30 seconds?

I was wondering how I could make the sprite object move 5 pixels every 0.2 seconds using dt every tick?

Yes! Exactly that, sprite.Y+(300/30*dt) will move 300 in 30 seconds

We can use the same logic for this second problem. The distance moved is (distance/time*dt), in this case 5/0.2*dt which is equal to 25*dt, so if it's still moving downwards we would use sprite.Y+(25*dt).

LevelUpJordan

OH, absolutely amazing,

I really appreciate your help and your explanation, thank you

No problem