Question: [![enter image description here](https://i.stack.imgur.com/qe9JA.png)](https://i.stack.imgur.com/qe9JA.png)

I know when it is centered at 0, Vpeak/sqrt(2) gives the RMS.

When I do FFT on LTspice, it gives a peak at -17dB which is equal to 0.141254

[![enter image description here](https://i.stack.imgur.com/0mP38.png)](https://i.stack.imgur.com/0mP38.png)

On the FFT page, program says that it shows RMS values. But, I don't think the RMS of this wave is 0.14 W. You are also welcome to answer for this question involving LTspice.

However, main question is : I actually don't think this question is specific to power calculation, anyway, **what is the method to calculate RMS of such sine waves?**

Answer: 
You want to know the rms of a waveform \$ f(t) \$.

There are two approaches here direct application of the formula:

$$RMS = \sqrt{\dfrac{1}{T} \cdot \int\_0^T (f(t))^2 \text{d}t}$$

Where T is the period of the signal. However there is another useful function:

$$RMS = \sqrt{RMS\_1^2 + RMS\_2^2}$$

Where \$RMS\_1\$ and \$RMS\_2\$ are the RMS values two signals that when added together give \$f(t)\$.

This only works provided the two signals do not share any frequency content in common. In this case a sine wave only contains one frequency as does DC (0 Hz which is different) so you can use this formula.

The RMS of a sin wave is \$ \dfrac{\sqrt{2}}{2} \cdot V\_{pk} = \dfrac{\sqrt{2}}{4} \cdot V\_{pkpk}\$ where \$V\_{pkpk}\$ is the peak to peak voltage. The RMS of a DC signal is the same as the DC value.
The RMS value you are seeing is a combination of the DC and AC components. Your signal is the sum of those two components and the RMS value is taken **of the sum**.

If you want only the RMS value of the AC portion, i.e., the sinusoidal portion, then using the SPICE controls to subtract the DC component and then take the RMS value of that result, which will now be only the AC portion of the signal.

Make sense?