Anybody know how to determine the closest orbits that an Earthlike planet could safely ride about a gas giant primary (such as Jupiter)
and the effect that orbital distance would have on  geologic activity and ocean tides?

I have seen them depicted in what looks like impossibly close proximities in movies and Discovery channel documentaries, but it seems to me by applying what little I know of the tidal equation, that those arrangements would yield devastatingly large ocean tides and land surface stresses.

thx

https://en.wikipedia.org/wiki/Roche_limit

Space Engine automatically calculates the Roche limit for all objects. You can see it at the bottom of the Orbital tab of the Info box:

If the orbit has 0 eccentricity and the moon is tidally locked to the planet (which it would definitely be), then there will be no tidal stress and no ocean tides (there will be a tidal bulge, but it will not move over time). There are only stresses if the orbit has any eccentricity (which all orbits do, to some extent). The more eccentric the orbit, the greater the tidal stress. But if you hypothetically have a perfectly circular orbit, then the planet will be fine as long as it's outside the Roche limit.

and the moon is tidally locked to the planet (which it would definitely be),

Oh No!  That ruins everything!   Is there any way to get around that?
While it's great news about dealing with the tidal surge, I can't have a 30 day long day on my planet....

Oh No!  That ruins everything!   Is there any way to get around that?

You just need to insert 'TidalLocked false' in your script. (Before the 'Surface' section.) Then you can give your Moon any rotation period.
But then there is again something like a tidal stress! In reality. In SE probably not.

I can't have a 30 day long day on my planet....

If your moon is as close to your planet as you initially indicated, its orbital period wouldn't be anywhere near 30 days. More like 20-30 hours.

The moon is located about 1.6 million miles away, at the edge of the parent planet's moon system in order to minimize the tidal forces. At this distance it has a 29 day orbit, and yet is still subject to 41 times the combined tidal forces that the Sun and Moon apply to the Earth here. Any farther away, and I've found that the parent star will distort the orbit and then strip the moon away from it's gas giant primary within a few hundred orbits.

Placing the moon super close to the primary might technically shorten the tidally locked day to within 20 or so hours, but it might violate the Roche Limit. Even if it doesn't, the primary would be so close and be so bright, completely filling the night-time sky that there would never be any darkness, not even close. Argueably, a more reflective planet like a cloud covered gas giant always filling the night-time sky would be even more blindingly bright than the sun would be on the sunlit side (square area of illuminance X brightness). Thus, no darkness or nighttime, ever, (except maybe an hour during eclipse time on the backside of the orbit), and a population of bleary-eyed humans trying to cope with perpetual overbearing daylight all the time. That would be hell on their circadian rhythms.

 So now you see my problem....

Placing the moon super close to the primary might technically shorten the tidally locked day to within 20 or so hours, but it might violate the Roche Limit.

You always can make the primary planet less massive.

Thus, no darkness or nighttime, ever, (except maybe an hour during eclipse time on the backside of the orbit)

How about the hemisphere faced away from the primary planet?

Placing the moon super close to the primary might technically shorten the tidally locked day to within 20 or so hours, but it might violate the Roche Limit.

You always can make the primary planet less massive.

Thus, no darkness or nighttime, ever, (except maybe an hour during eclipse time on the backside of the orbit)

How about the hemisphere faced away from the primary planet?

Good point, hadn't thought of that.... but then you lose the benefit of placing it around a gas giant in the first place (seeing it decorate the sky!)....
You would have to board a plane and fly 20 hours to see the main attraction, or live on the side of perpetual daylight.

No matter what, it seems that you lose the depicted worlds of Avatar and all those sci-fi movies and documentaries showing these beautiful worlds with big Jupiters/Saturns in the sky.

God, nature is a biyotch!

What is the relationship between tidal forces and the rate of rotational braking down to a tidal lock?

In simple terms is it, say, a linear relationship, a root relationship, or by the square?

Along the same lines, what is the relation between ocean tidal height and tidal force, linear, root, or square?

What is the relationship between tidal forces and the rate of rotational braking down to a tidal lock?

T is a tidal braking time (in billions of years) of the satellite, from starting rotation period p₀ (days) to final p (days),
a - semimajor axis in AU,
j - inertia moment of the satellite (typical is 0.3 - 0.4),
Q/k is a tidal dissipation speed, it is known for Earth (500) and Mars (700).

If you want to calculate full tidal synchronization time, set p = orbital period.

Along the same lines, what is the relation between ocean tidal height and tidal force, linear, root, or square?

h is the tidal wave height in open water,
h_earth is the tidal wave height on Earth (30 cm),
a in AU.

Arhg! I can't make it work! I don't know how to use that equation! I keep getting these rediculously large answers,
like wave heights of millions of miles high. I just don't know how to use the a, Mprime, Msat, Rsat and Rearth terms.

Let me try asking this way;

There are two planets; "Planet A" and "Planet A (perfect copy)", which is a perfect copy of Planet A in every way.

A man is standing on the beach of "Planet A" and observes the ocean tide height to be 1 meter over the course of a day.
 That Planet is subject to tidal force 1X.

Another man is standing on the corresponding beach on "Planet A (perfect copy)". That planet is subject to the tidal force 10x.
 What will be the tide height that this second man sees?

Thank you, and I'm sorry for being so dim here....

senuaafo, the tide height in open water is proportional to the tidal force, so in your example it would be 10x higher.  However, the tide interacts with and sloshes around landmasses in a complicated way.  So the actual tidal variation a person sees on a beach can be very different.  It might not be accurate that the beach tide would also increase by a factor of 10.

Another general fact with the tidal force is that it is inversely proportional to the cube of the distance to the satellite, as opposed to the inverse square law for the gravitational force.  This is because the tidal force represents the change in strength (and direction) of the gravitational force from one point to another.

Thank you. That's is the type of answer I was looking for..... linear 1:1 in open water, while beaches and bays complicate things.