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I currently having some problems with calculating the number of sample I have after I've sampled my audio file.

The audio file is divided into frames, and each frames are then sampled.

Framing is like this.

enter image description here

A frame constant in length => my frame length is $25 \text{ ms}$ and between each frame will there be an overlap meaning some of the data will be reused. The overlap in my situation is $10 \text{ ms}$.

I tried sampling an audio file of length $1 \text{ sec}$, which normally should have given $16000$ sample points, but I am getting $25186$ sample points?

I then tried to include the overlap, which would occur for each $\frac{16000}{400} = 40$ times, and each overlap will create an extra delay of 10 ms which in total end up being $0.4 \text{ s}$. So the full length would be $1.4 \text{ s}$ => $16000\cdot 1.4 = 22400$ samples, which is close but not the actual number..

So my question here is, is there anything wrong with my calculations, or is it correct and the overlap creates $0.4$ seconds extra data, and that would in total create $22400$ datapoint, meaning that either the framelength, frameoverlap or samplerate is wrong?..

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    Should be 67 25ms intervals. Unfortunately that gives 375.9 samples per interval, which is not a nice number at all. – 2017-01-11
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    how dare you... calling rational numbers not nice... – 2017-01-11
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    Further thoughts: this number is 98 * 257. That's pretty close to having a frame skip of 10ms, as opposed to an overlap of same. That suggests a sample rate of 10240Hz, assuming each frame gets both fenceposts, not 16000Hz. How are you determining sample rate? – 2017-01-11
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    The software i am using uses 16000Hz as standard, and as I haven't changed that i guessed that would be the case. – 2017-01-11

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