Can a Sinewave Be Generated from an Oval Phasor?

[tomelex]

I am not an expert in math, but as an electronics tech we use sinewaves all the time. I understand how it is derived from a circle well enough ( a phasor rotating about a circle).

And that makes sense as that is how electricity can be generated due to the orientation of the generator physical parts.

However, let's say we are using an oval instead of a circle with our phasor, can we still technically say it is a sinewave, or do we say it is sinusodal, or what?

I think that by definition below we can call rotating about an oval a sinewave, but then I don't think the numbers would be right, as official sine numbers (sine of 90 is1) are only for a circle , right?

My old Radio Shack dictionary of electronics states:

SINEWAVE: A wave which can be expressed as the sine of a linear function of time, space, or both.

if sine is opposite divided by hypotenuse, then to me it appears that a phasor through an oval will get you a sinewave? I am confused.

Simply, does it have to be a circle to be a sinewave? Is there an official sinewave and anything else is not?


Thanks

Tomelex

 

[sachinism]

well my answer to this would be

yes and no

no in the sense that the same rules owuldnt follow as u said and we would have to develop a altogether new system in such a case

yes int he sense that this owuld be perfectly true if it ould be giving us the same sine values for all the angles as in the general conventional sense but for this we would have to have a corresponding analogue of pythogoras theorem .

 

[tomelex]

Thank you sachinism for your reply.

given the definition of a sinewave in the previous post, then I assume that a linear function means say degrees, seconds, etc.

So, a sinewave as I learned in electronics, developed from a phasor in a circle, is not the only wave that can be described as a sinewave, in fact, any waveform that can be sort of divided in time or space into increments, and then you do the opposite over the hypotenuse thing, then you have a sinewave.

So, sinewave just means you did opposite divided by hypotenuse, not that it was a circle to start with at all. The sine function on my calculator assumes you did a circle.

Therefore, when you invoke the sine function on your calculator, if you are not measuring a circle, you will not be 100% accurate, but if the waveform looks sort of like a circle, then pressing the sine button on the caluclator will get you close. I think.