I'm curious if someone could give me a hint as to how external forces are calculated where FEM solid objects are concerned.

Take a simple solid object. By default a Gravity DOP will be attached, adding to the downward force. This will work as expected.

Now replace the Gravity DOP with an Uniform Force DOP or any other node that modifies force. Notice how they barely influence the object at all. To get a behaviour similar to the Gravity DOP you need to input forces of -1000 or more!

Why is Gravity “special” here?
Is there a drag parameter that is set so high by default that other forces are almost ignored, and if so, where is it hidden?

Cheers
Michael

The responsible parameter is the “Mass Density” on the Solid Object.
This is set to 1000 by default (1000kg per cubic metre).
Newton's second law, F = M a, is the reason that you're not seeing much effect from the Uniform Force with small values.
The mass M in this equation is related to the Mass Density parameter.

The “Force” parameter in the Gravity Force DOP actually specifies the acceleration that the force would cause, not the magnitude of the force itself.
To make a Uniform Force that matches gravity, you need to mulitply the acceleration (default -9.80665) by the mass density (default 1000), following F = m a.
So a Uniform Force that matches the default gravity would have its Force parameter to be set to (0, -9806.65, 0).

I hope this clarifies things a bit.

Michiel

Thanks, that makes sense
(though it'd probably be good if the naming on the gravity node would be slightly different than on the other force nodes, since calling it all “Force” is pretty confusing tbh.)

ripclaw
since calling it all “Force” is pretty confusing tbh.)

This also confused another person recently, so, perhaps ‘Fictitious Force’ might help disambiguate it:
https://en.wikipedia.org/wiki/Fictitious_force#Gravity_as_a_fictitious_force [en.wikipedia.org]

wiki
All fictitious forces are proportional to the mass of the object upon which they act, which is also true for gravity.