Say you draw 1700w due to inverter efficiency, which is 142Amps. 230Ah capacity is usually at 0.01C which means this is the capacity at (Battery Ah)*0.1 Amps Draw - which is 23Amps in your case. If you go to 0.5C (115A) for example, your capacity is around 70%. After understanding this, there are a few more things to be aware of:
- On such amperage you are going to create a lot of heat, I don't think you want to run on such amperage for a long time. (Rule of thumb - don't let your battery get to 50C). Batteries that get warm while operating/in storage die very early
- Lead Acid batteries should not be discharged to more than 50-70% (depending of their technology and quality) of their full capacity - on your case it means DO NOT discharge more than 115Ah! (This is called DoD, discharging more than this will shorten the life of your battery drastically)
- This amperage requirs very thick cables, consider going to higher voltage batteries (in series) and inverter to lower the amperage
Your mistake is treating amp-hours (Ah) as energy. They are not. Amp-hours are units of charge. The efficiency of the inverter applies to the power, not the current:
$$\rm output\ power = input\ power \times efficiency$$
So if you have a 1500 watt load, your average input power would be:
$$\frac{1500 \mathrm W}{85\%} \approx 1765 \mathrm W$$
Which means your average input current would be:
$$\frac{1765 \mathrm W}{12 \mathrm V} \approx 147 \mathrm A$$
Which gives a run time of:
$$\frac{230 \mathrm{Ah}} {147 \mathrm A} = 1.56\ \mathrm {hours}$$
Note that 147 amps is a dangerously large current, so be sure your battery is wired to the inverter properly.