Re: Accepts Lights+Shadows via expression or Scrip...

Dan Ebberts · ‎Mar 19, 2010

Something like this should get you headed in the right direction: { function shadowsOn(){ var myLayers = app.project.activeItem.selectedLayers; for (var i = 0; i < myLayers.length; i++){ myLayers.acceptsShadows.setValue(true); } } function shadowsOff(){ var myLayers = app.project.activeItem.selectedLayers; for (var i = 0; i < myLayers.length; i++){ myLayers.acceptsShadows.setValue(false); } } function lightsOn(){ var myLayers = app.project.activeItem.selectedLayers; for (var i = 0; i < myLayers.length; i++){ myLayers.acceptsLights.setValue(true); } } function lightsOff(){ var myLayers = app.project.activeItem.selectedLayers; for (var i = 0; i < myLayers.length; i++){ myLayers.acceptsLights.setValue(false); } } function createUI(thisObj) { var myPanel = (thisObj instanceof Panel) ? thisObj : new Window("palette", "My Tools", [100, 100, 300, 300]); myPanel.shadowsOnBtn = myPanel.add("button", [10, 10, 150, 30], "Accept Shadows on"); myPanel.shadowsOffBtn = myPanel.add("button", [10, 50, 150, 70], "Accept Shadows off"); myPanel.lightsOnBtn = myPanel.add("button", [10, 90, 150, 110], "Accept Lights on"); myPanel.lightsOffBtn = myPanel.add("button", [10, 130, 150, 150], "Accept Lights off"); myPanel.shadowsOnBtn.onClick = shadowsOn; myPanel.shadowsOffBtn.onClick = shadowsOff; myPanel.lightsOnBtn.onClick = lightsOn; myPanel.lightsOffBtn.onClick = lightsOff; return myPanel; } var myToolsPanel = createUI(this); myToolsPanel.show(); } Dan

Dan Ebberts · ‎Mar 18, 2010

This simple example script turns off accept lights and shadows for selected layers (it assumes those properties haven't been keyframed): { var myLayers = app.project.activeItem.selectedLayers; for (var i = 0; i < myLayers.length; i++){ myLayers.acceptsShadows.setValue(false); myLayers.acceptsLights.setValue(false); } } Dan

Dan Ebberts · ‎Mar 09, 2010

This should give you what you're looking for: Math.floor(random(8))*45 I didn't include 360 because that's the same as 0, but if you want it, just change the 8 to a 9. Dan

Dan Ebberts · ‎Feb 16, 2010

Time remapping doesn't seem to update in the Final_Comp the way you would expect. How about scrapping the time remapping idea altogether and use layer markers instead? Apply the layer markers to the All_Layers layer in the Final_Comp and set the comment of the marker to the index (layer number) of the layer that you want to appear at that point. Apply this expression to each layer's (in the All_Layers comp) opacity: L = comp("Final_Comp").layer("All_Layers"); n = 0; if (L.marker.numKeys > 0){ n = L.marker.nearestKey(time).index; if (L.marker.key(n).time > time) n--; } if (n > 0){ idx = parseInt(L.marker.key(n).comment,10); if (idx == index) 100 else 0; }else{ 0 } Dan

Dan Ebberts · ‎Feb 16, 2010

Ah, so you want a time value of 1 frame to turn on Layer 1, etc. The expression I gave you assumes a time value of 1 second. Here's it is modified for frames: L = comp("Final_Comp").layer("All_Layers"); if (timeToFrames(L.timeRemap) == index) 100 else 0 Dan

Dan Ebberts · ‎Feb 09, 2010

Your original expression doesn't work mainly because an expression can't use valueAtTime() to get a value that it calculated previously. The only way an expression can retrieve a previous result for the property to which it is applied is to recalculate it. You can't even apply valueAtTime() to anything other than a property, which is why you got the error message when you tried to apply it to a variable. Dan

Dan Ebberts · ‎Feb 09, 2010

I just tried it on a 20 second comp (at 29.97 and 25 fps) and I don't get any repeated frames. I guess I'd need to see your project file to figure out why it doesn't work for you. Dan

Dan Ebberts · ‎Feb 09, 2010

Hmmm.... I don't know why it's not working for you. I just tried it and I don't get any repeated frames as far out as I checked (200+ frames or so). Dan

Dan Ebberts · ‎Feb 09, 2010

Your English is fine. Maybe it's me that's not articulating very well. 🙂 The expression I posted is a time remapping expression that should do what you want. Did you try it? Dan

Dan Ebberts · ‎Feb 09, 2010

To avoid generating the same number twice in a row, you need to keep track of what's happened in the past. How many different images do you have? This expression assumes 10, but just edit the first line to match your number. This is one of those expressions that has to recalculate what's happened on each previous frame, so it might bog down if your comp is fairly long. nFrames = 10; seedRandom(index,true); n = Math.floor(random(nFrames)); f = timeToFrames(); for (i = 0; i < f; i++){ n += 1 +Math.floor(random(nFrames-1)); n %= nFrames; } framesToTime(n) Dan

Dan Ebberts · ‎Dec 24, 2009

Nevermind. I discovered that I was using PF_New_World instead of PF_NewWorld. Yikes. Dan

Dan Ebberts · ‎Dec 24, 2009

I've got an inputLayer_param that I want to stretch to fit my input_world. In 8bpc, I'd just create a new input_world-sized PF_EffectWorld and use PF_Copy. I can't figure out how to do it in 32bpc, because I can't create a PF_EffectWorld of the correct bit depth. Maybe I'm barking up the wrong tree. Any guidance would be appreciated. Dan

Dan Ebberts · ‎Dec 17, 2009

This should work: angle = 45; w = wiggle(3,100) - value; a = degreesToRadians(angle); x = w[0]*Math.cos(a); y = w[0]*Math.sin(a); value + [x,y] Dan

Dan Ebberts · ‎Dec 14, 2009

To get loopOut() to cycle the way you expect, you just need to make sure you set your keyframes up so that the last keyframe has the same value as the first. So in your last example, try moving your second 0 keyframe to frame 99 and setting a keyframe for -222 at frame 100. Dan

Dan Ebberts · ‎Nov 25, 2009

It just converts the null and camera postions to world space coordinates and then calculates the distance between them with length(). Dan

Dan Ebberts · ‎Nov 24, 2009

Try this: L = thisComp.layer("Cam_control"); length( L.toWorld(L.anchorPoint), toWorld([0,0,0])) Dan

Dan Ebberts · ‎Nov 14, 2009

Thanks for chiming in Bruce. My original objective was to have a new set of my effect's perlin noise elements come up if the user applied more than one instance of the effect to the same layer (I already incorporate the layer index so that each layer gets seeded differently). I figured using the effect index would leave the original instance untouched (unless the user moved it around in the effect stack) but additional instances would get a different seed. It seemed like a simple idea. I eventually just gave up on it, figuring that it's unlikely that this particular plugin will get applied more than once to the same layer, but I'd still like to know the best way to do if for future reference. Dan

Dan Ebberts · ‎Nov 05, 2009

You can't access one expression from another, but you could construct your expression in a text layer and the evaluate that from your other expressions. So, you could put something like this a text layer: x = 100; y = 100; [x,y] And then put something like this in another layer's position expression: eval(thisComp.layer("Text Layer 1").text.sourceText.value); Dan

Dan Ebberts · ‎Oct 22, 2009

Not directly, but there may be a work-around, depending on what you're trying to do. If you paste the path into the position property of a null, AE creates a smooth 2-second animation. You can like to the null's position and use a positon expression with the light (using valueAtTime()) to traverse the path on any timescale you want - stop, back up, etc. If what you're looking for is a live link between the path and the light though, then this doesn't help. A script has access to all the mask stuff, so depending on exactly what you're trying to do, you could use a little helper script to update your light position keyframes after you edit the path. Dan

Dan Ebberts · ‎Oct 21, 2009

Yes, that's correct. Dan

Dan Ebberts · ‎Oct 21, 2009

That last position expression doesn't work quite right if the layers are parented to the ring. This should be better: offset = -50; n = thisComp.numLayers - 2; angle = index*Math.PI*2/n; if (hasParent){ center =[parent.width/2,parent.height/2,0]; radius = parent.width/2 + offset; center + [radius*Math.cos(angle),radius*Math.sin(angle),-height/2] }else{ ringLayer = thisComp.layer("ring"); center = ringLayer.transform.position; radius = ringLayer.width/2 + offset; center + [radius*Math.cos(angle),-height/2,radius*Math.sin(angle)] } Dan

Dan Ebberts · ‎Oct 21, 2009

For 3D, the position expression would look like this (now assuming there is a camera in the layer stack below the layers to be distributed): if (hasParent){ center =[parent.width/2,parent.height/2,0]; radius = parent.width/2; }else{ ringLayer = thisComp.layer("ring"); center = ringLayer.transform.position; radius = ringLayer.width/2; } offset = -50; radius += offset n = thisComp.numLayers - 2; angle = index*Math.PI*2/n; center + [radius*Math.cos(angle),-height/2,radius*Math.sin(angle)] And y rotation like this: n = thisComp.numLayers - 2; offset = 90; offset - index*360/n; Dan

Dan Ebberts · ‎Oct 21, 2009

It occured to me that you might want to parent the layers to the ring. If so, you'd need to change the position expression to something like this (which will handle both the parented and unparented cases): if (hasParent){ center =[parent.width,parent.height]/2; radius = parent.width/2; }else{ ringLayer = thisComp.layer("ring"); center = ringLayer.transform.position; radius = ringLayer.width/2; } offset = -50; radius += offset n = thisComp.numLayers - 1; angle = index*Math.PI*2/n; center + radius*[Math.cos(angle),Math.sin(angle)] Dan

Dan Ebberts · ‎Oct 21, 2009

Assuming that your ring layer is at the bottom of your layer stack and is the only other layer in the comp besides the layers to be distributed (I'm also assuming it is named "ring"), you can distribute your layers with a position expression like this: ringLayer = thisComp.layer("ring"); offset = -50; radius = ringLayer.width/2 + offset; center = ringLayer.transform.position; n = thisComp.numLayers - 1; angle = index*Math.PI*2/n; center + radius*[Math.cos(angle),Math.sin(angle)] Adjust the "offset" parameter (or connect it to a slider) to adjust the distance from each layer to the center of the ring. A rotation expression like this should handle orienting each layer toward the center of the ring: n = thisComp.numLayers - 1; offset = 180; index*360/n + offset; Adjust the "offset" parameter as necessary (or connect to an angle control) to get the layers to point exactly where you want them to. Dan

Dan Ebberts · ‎Oct 17, 2009

Yikes. I was afraid of that. Thanks for the info though! I just wanted to use it as a seed (combined with the layer ID) to make sure each instance of my effect on any given layer would generate different random numbers. In my case though, it's unlikely that anyone would apply the effect more than once to the same layer, so I think I'll skip it for now and just use the layer ID. It would be useful in the future though, to have an easy way to do it. I'm going to leave the topic as "not answered" just in case anyone else has any ideas. Thanks again. Dan

Dan Ebberts · ‎Oct 16, 2009

I know this must be simple, but I'm not finding it. How does an effect get its own effect_index? Dan

Dan Ebberts · ‎Oct 12, 2009

Try setting the variable "factor" to a smaller number. Dan

Dan Ebberts · ‎Oct 12, 2009

Using an expression to control the speed of something based on the speed of something else is always a bit of tricky business. What you really need in this case, I think, is to link the elapsed time-remapped time of the QT movie to the total distance traveled by your other layer. That requires a frame-by-frame integration operation. Since it has to look at each previous frame, it can bog down as the length of your piece increases. Here's how it would look though: L = thisComp.layer("Layer 1"); // your speed layer factor = 25; // increase this to increase overall speed f = timeToFrames(); fCount = 0; accum = 0; while (fCount < f){ accum += L.transform.position.speedAtTime(framesToTime(fCount))*thisComp.frameDuration; fCount++; } accum*(factor/1000) Dan

Dan Ebberts · ‎Oct 09, 2009

Actually, your random expression would look like this: random([0,10,50],[100,50,80]) giving you an x range of 0 to 100, y of 10 to 50, and z of 50 to 80. wiggle() generates values of the same dimension as the property to which it is applied. random() takes one or two parameters (scalars or arrays). If you only provide one parameter, the other is assumed to be zero, (or [0,0], or [0,0,0], or [0,0,0,0], etc.) and generates values with dimensions matching the parameters that you supply. Dan

Dan Ebberts · ‎Aug 27, 2009

Thank you so much! Dan

Dan Ebberts

Dan Ebberts

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