[Darby Edelen]

The equation that Walter posted previously:

zoom = thisComp.width/(2*Math.tan(degreesToRadians(FOV/2)));

Can be solved for FOV like this (assume ‘zoom’ has been assigned the zoom value of the camera):


zoom * 2 * Math.tan(degreesToRadians(FOV/2)) = thisComp.width;

Math.tan(degreesToRadians(FOV/2)) = thisComp.width / (2 * zoom);

degreesToRadians(FOV/2) = Math.atan2(thisComp.width, 2 * zoom);

FOV/2 = radiansToDegrees(Math.atan2(thisComp.width, 2 * zoom));
FOV = 2 * radiansToDegrees(Math.atan2(thisComp.width, 2 * zoom));


Since we are using thisComp.width we are solving for the horizontal FOV.

However, what Walter keeps coming back to is that converting the camera zoom value into a FOV is likely not necessary as it will probably end up being converted back (perhaps even unknowingly) into a zoom value at some point. I’ve seen expressions that have 10 extra lines and 20 extra operations all to do unnecessary conversions from zoom to FOV back to zoom or worse.

The only likely reason I can see for having any conversion from FOV to zoom or vice versa is for allowing FOV input from/output to a user. Otherwise all calculations should probably be done with the zoom value.

To answer your original question the ‘default’ camera (the one that exists if you don’t have a camera in the composition) has a horizontal FOV of about 39.6°. I’d recommend using the default zoom value though, which you can find this way:

defaultCameraZoom = thisComp.width / 0.72;

0.72 is the ratio of the default camera film back (36mm) divided by the default camera focal length (50mm). The triangle formed by the zoom and composition width values is a projection of the triangle formed by the film back and focal length. The ratios of the two triangles are the same.

This ratio only necessarily applies to the default camera (36mm film back / 50mm focal length).

Darby Edelen

[Darby Edelen]

[Brian Jones] “well, not evil but tricky – lots of pre-planning.”

Nope, just evil 🙂

Modelling works best with lots of pre-planning either way.

Darby Edelen

[Darby Edelen]

Booleans are evil.

Since this is radially symmetrical the way I’d do this is by sub-d modelling a very small section of the circle, say from one ridge to the next, and then cloning that geometry using a Radial cloner to create the full circle.

Take a look at this video for some inspiration: https://vimeo.com/57225673

Darby Edelen

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[Darby Edelen]

This is not an easy undertaking even if you’re familiar with Cinema 4D.

Getting this sort of smokey render is challenging without spending some money on a 3rd party plug-in like Turbulence FD or X-Particles. If you monkey with MoGraph and the free Curl Noise effector (https://github.com/CreativeTools/ct-curl-noise) you might be able to get some decent looking curly particles but rendering them will be challenging. Cinema 4D doesn’t have great built-in tools for rendering smoke. PyroCluster or Hair might work.

Basically there are a lot of distinct tool sets to learn and not all of them are available to every user (Hair, for example, is only in Studio I believe).

Darby Edelen

[Darby Edelen]

As far as I understand it, Realflow is pretty industry standard. However, Houdini has some pretty awesome fluid tools in it as well. The learning curve may be steeper with Houdini, but it can do much much more than just fluids. There’s also a learning edition available for free so you can test drive it.

Darby Edelen

[Darby Edelen]

As Roland has already suggested the image you’ve provided is impossible to create with one path. The smallest number of paths you could possibly use to create that (or something very close to it) is 2. A path cannot branch into two directions as it would have to where the body of your snail meets the shell.

Additionally, the Unite operation needs closed paths to work properly so that Illustrator knows what is ‘inside’ the path versus ‘outside’ the path.

Darby Edelen

[Darby Edelen]

[Roland R. Kahlenberg] “I didn’t think that interpreting the Alpha correctly was a solution – wouldn’t the pre-mult alpha contain some RGB pixels of the rendered comp?”

You probably know this already, but in the interest of other readers I’ll try to be comprehensive 🙂

A Straight and Premultiplied render will both have the exact same alpha. The difference is in the RGB.

If you render straight then the RGB channels extend beyond the edges of the alpha channel as they will be masked by the alpha in the composite.

If you render premultiplied then these ‘extended’ straight RGB channels are multiplied by the alpha. The fully opaque areas are multiplied by white so they look the same. The transparent areas are multiplied by black (or some other arbitrary color, but usually black) and the semi-transparent areas become a mix of black and the original RGB values.

If you interpret premultiplied footage correctly then AE will unpremultiply (divide) the RGB values by the alpha and you’ll end up with the original RGB values, the same ones you’d get from a straight alpha, since division (or ‘un-multiplication’… glad they didn’t call it that in grade school) is the inverse of multiplication.

I’m of the opinion that neither straight or premultiplied is necessarily the better option, the workflow just needs to match the chosen method.

Darby Edelen

[Darby Edelen]

You should rotate 180°. If you flip on the Y-axis you’ll get behavior like a mirror, right and left will be switched.

Darby Edelen

[Darby Edelen]

Import the source R3D files and then set the ProRes files as proxies. This will work best if the files are the same aspect ratio (both 16:9 for example).

https://helpx.adobe.com/en/after-effects/using/footage-items.html#placeholders_and_proxies

You can then work on the ProRes footage and render using the R3D footage.

Darby Edelen

[Darby Edelen]

[xavier corral] “temp = motionTracker(“Tracker 7”)(“Track Point 1″).attachPoint; [temp(0)+15,temp(1)];”

Try this instead:

temp = motionTracker("Tracker 7")("Track Point 1").attachPoint;
temp + [15,0];

Darby Edelen