using a 1000 watt inverter with a Gen 2

Yes you can connect it to jump start point. Ground would be better connected to the body. Nuts that hold the front strut to the body would be one good and easy to access location. 1000w is pretty much the maximum that you can use so don’t use full power for long time.

 

There are lengthy past threads as that was a popular thing to do for elec outages etc.

 

As stated, there are many threads about power inverters that go into depth on this topic. Definitely don't connect to the jump point if you're looking to draw that much power. A couple hundred watts from the jump point is okay, but for a thousand watts, you definitely want to connect directly to the battery in the trunk with a fuse. The Prius can provide around 60A at ~14V from the inverter continuously. More than than that and you'll be draining from the battery.

 

There should be no problem connecting the 1000w inverter to jump start point. Let’s say the voltage would drop to maybe 11.9V and current would be 110A and DC-DC converter could only deliver 60A (DC-DC converter should actually be able to deliver more current as the voltage drops). Even then most of the current/power would be coming from DC-DC converter. Rest of the current 110A-60A=50A would be coming from the 12V battery at the back through relatively thick wire and 120A fuse.

Connecting directly to 12V battery won’t really make it much or any better as the voltage of the battery could actually drop even lower as Prius 12V battery has relatively high internal resistance (especially if it’s not new) and if connecting the load to the rear more of the current would go through that long wire that feeds the 12V battery.

 

I would connect to the battery to minimize the chance of damaging the inverter. I also would not want any chance of blowing the main 120A battery fuse in case of problems. There my preference is definitely to tap off the battery instead of the jump point.

The 1000W inverter at maximum capacity is probably 75% efficient, which means a draw of ~1300W. If the voltage drops down to 12V, you're looking at ~110A, which is very close to blowing that 120A fuse. I would therefore tap off the battery with a 120 or 140A fuse to not risk disabling the car's electronics.

 

If you keep the Prius powered on then only the current that cannot be supplied by DC-DC converter is going through that 120A fuse. So less than 50A will be going through the 120A fuse. That’s less than half what the fuse is rated for so there are no problems.

Of course if for some reason you had to use the inverter without Prius being powered on then you definitely want to connect it directly to the 12V battery. But you shouldn’t use the inverter without Prius being powered on anyways as that would quickly drain the 12V battery.

 

Does anyone know what kind of safety features the DC-DC converter has?

I find the wiring to the battery to be fairly small (about 8Ga or under 10mm^2 if I had to guess).

I am running an amplifier from the battery and want to add another one. I have considered running the second amplifier from the converter/jump-point and also connect it to the first amplifier. Basically parallel the existing 10mm^2(?) main positive lead between DC-DC-converter and the battery with another >10mm^2 wire, fused at both ends. This would increase and stabilise the voltage at the amplifiers greatly.

If the 120A fuse under the bonnet is a safety feature for the DC-converter, then this is obviously a stupid idea...

As far as I know, the white (?) wire at the battery monitors the batteryvoltage and is a feed-back loop to the DC-DC converter to adjust for the voltage drop through the positive PLUS wire to the battery.
So imho, hooking it (amplifier, 1000W inverter, etc) up to the jump-point would indeed limit the current through the fairly long positive 12V wire and therefor reduce losses and limit the total draw.

Useless exercise as it is all guesswork...: suppose the positive wire to the battery is 5 meters long and indeed just 8Ga.
8Ga is about 0.002ohm per meter, so 0.01 ohm for 5 meters. Ignoring the resistance of the chassis, with a 100A current, we loose one volt in this PLUS battery wire (Voltage is Current X Resistance). And since Power = Voltage X Current = 1Volt X 100A = 100W EXTRA which is both lost in the wire AND needs to be converted by the converter (in addition to the ~1400W it has to convert.)
If we assume the loss in the chassis is equal, then double this powerloss...