[Solved]Variable duration of animation

DRAX

Here is another idea:
You could define PropertyChange on PropertyAnimation to change duration depending on state.
Example:

@import QtQuick 1.1

Rectangle {

width: 360

height: 360

property bool shown: true

Rectangle {

    id: rect

    width: parent.width

    height: parent.height

    color: "red"



    /*Behavior on y {

             NumberAnimation { duration: 2000 }

         }*/

}



states: [

State {

        name: "show"

        when: shown == true

        PropertyChanges { target: rect; y: 0}

        PropertyChanges { target: animation; duration: 2000}

    },

State {

        name: "hide"

        when: shown == false

        PropertyChanges { target: rect; y: rect.height }

        PropertyChanges { target: animation; duration: 1000}

    }



]

transitions: [

    Transition {

        from: "*"

        to: "*"

        reversible: true

        PropertyAnimation {

            id: animation

            target: rect

            properties: "y"

            duration: 2000

        }

    }



]

MouseArea {

    anchors.fill: parent

    onClicked: {

        shown = !shown

        print(shown)

    }

}

}@

riddle

You misunderstood me. Maybe I wrote it unclearly.

duration: (shown ? 1000 : 2000) does not solve anything.
The issue is that when you click once (animation starts now), and you click once again. Then the movement is short, but the duration is still 2000s. No matter how long the distance is.
Please run and check how it works..

Simply, I want to have constant speed no matter how many pixels are left to finish the animation.

sierdzio

[quote author="riddle" date="1355145150"]Simply, I want to have constant speed no matter how many pixels are left to finish the animation.
[/quote]

Use physics, then. time = distance/velocity. In your case, y should point to the current value when you invoke it in duration. Or you can temporarily stop the animation, get y and then use it to calculate duration.

riddle

bq. Use physics

Sure, but I have problem with using property y in binding calulcations. I get binding loop.

DRAX

I agree with sierdzio, you must use Y then.
No need for bounding.
You can do it on this way:

when you click on button, grab current position (A) and position where it should go (B)
subtract those two and take absolute value (abs(A-B))
And you are ready to go.

riddle

Maths says
When going from show to hide:
@duration: (rect.height-rect.y)/rect.height2000@
and when going from hide to show:
@duration: rect.y/rect.height2000@

But it doesn't work... we have QML PropertyAnimation: Binding loop detected for property "duration".
What's going on?

DRAX

Well, problem is that you are using binding.
Don't use it!
As I said above, when you click on button (rectangle) set formula (calculate, not bind).

Like this:
@
MouseArea {
anchors.fill: parent

    onClicked: {
        shown = !shown
        print(shown)
        animation.duration = (shown ? rect.y/rect.height*2000 : (rect.height-rect.y)/rect.height*2000)
    }
}

@

riddle

So there is no declarative way to do so?
Maybe it's a good idea to add constVelocity boolean property to animations. It could ensure that when transition is reversible, no matter when you reverse the animation, it will play with constant speed.

Oh, and one more thing. Could someone explain me why it does not work with properties (duration: (rect.height-rect.y)/rect.height*2000) ?

DRAX

In declarative way, it is not possible, only via JavaScript.

Duration cannot be bound to Y since it is one which changes Y.
Code that you posted should work like this:

animation started and changes Y
animation duration depends on Y so it updates
animation instance either resets or creates another one which changes Y (and from here you can see the loop)

In other words:
-change Y (animate)
-Y is changed, update duration (bound to Y)
-duration is changed, change Y (animate)
-Y is changed, update duration
-duration is changed, change Y
-Y is changed, update duration
-duration is changed, change Y
-Y is changed, update duration
...

riddle

Thank you for explanation.

DRAX

Anytime :)