#### Scottywatt

##### Member

- Location
- Mass

- Occupation
- Electrician/ Construction

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- Thread starter Scottywatt
- Start date

- Location
- Mass

- Occupation
- Electrician/ Construction

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- Location
- Mass

- Occupation
- Electrician/ Construction

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- Location
- Springfield, MA, USA

- Occupation
- Electric motor research

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Very close.

You understand DC voltage; a steady voltage from something like a battery.

For an AC voltage, RMS is the 'effective voltage' in the sense that it would supply the same power to a resistor as the DC voltage.

The calculation is a bit more complicated than a simple average cutting off the peaks, and many modern meters do the proper calculation internally.

The reason that it is important to know that it isn't a simple average is because some older meters would take the simple average and then scale the result, giving an incorrect reading if you don't have a true sine wave.

-Jon

- Location
- Mass

- Occupation
- Electrician/ Construction

Ok

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- Location
- Mass

- Occupation
- Electrician/ Construction

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- Location
- Massachusetts

You are correct. Hertz or Hz is a shorthand name for "cycles per second", which measures frequency (f).

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Period (T) is the reciprocal of the frequency, and has the units of seconds. Reciprocal means T = 1/f, and f = 1/T. The period is the time to complete a full cycle. I.e. the time between equivalent points on two consecutive cycles of a function that repeats itself periodically as a function of time. You can measure from crest-to-crest, you can measure from trough-to-trough, you can measure from ascending node to ascending node, or you can measure from descending node to descending node. What is essential is that you measure it between equivalent points on consecutive cycles. It has a lot in common with wavelength in that is is measured from equivalent point to equivalent point, but the essential difference is that period is term for time, and wavelength is a term for distance.

A frequency of 60 Hz doesn't necessarily mean that the generator rotor spins at 60 rotations per second (i.e. 3600 rpm). It could be the case that you have a 3600 rpm generator and a 60 Hz AC waveform, however different configurations of the magnets and coils can enable a generator to rotate at another speed. Such as 1800 rpm, which is the equivalent of 30 Hz. Usually there is a whole number or unit fraction ratio between generator rotation rate and electrical frequency.

- Location
- Mass

- Occupation
- Electrician/ Construction

So the half way point would be 0.0055555

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- Location
- Massachusetts

So the half way point would be 0.0055555

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Depends on what you mean by half way point. Half way through a cycle is 180 degrees, or half the period. Voltage being at half its maximum would occur at 30 degrees and 150 degrees in a cycle, which are special angles whose sine is equal to exactly 1/2.

You can translate 30 degrees and 150 degrees to seconds, by setting up a ratio with the period on top and 360 degreeson the bottom.

- Location
- Mass

- Occupation
- Electrician/ Construction

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- Location
- Mass

- Occupation
- Electrician/ Construction

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- Location
- Massachusetts

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Is there any part of what I said, that I could rephrase for you?

I'd recommend you look up the the basic vocabulary of waveforms, and master these terms. There are Wiki articles on each of them that elaborate in further detail, with great visuals.

Amplitude

Peak-to-peak amplitude

Period

Frequency

Crest & trough

Phase offset

Special right triangle trig ratios

And a sine wave. Like 60Hz. Or 440Hz middle A on the piano.Is there any part of what I said, that I could rephrase for you?

I'd recommend you look up the the basic vocabulary of waveforms, and master these terms. There are Wiki articles on each of them that elaborate in further detail, with great visuals.

Amplitude

Peak-to-peak amplitude

Period

Frequency

Crest & trough

Phase offset

Special right triangle trig ratios

I'll get me coat.............

- Location
- Ann Arbor, Michigan

Scottywatt:

The title of your thread is :

"Hi everyone I'm looking how to find time of a sine wave"

In retrospect I believe this is the question you were ask in your test. But it is hidden int the title and not explicitly stated as being the exact question. However, it is not a very complete or accurate question. What is the meaning of "time of a sine wave" ? Does that mean the period of the sine wave? I doubt you ever had a high school class in physics where one would be taught the formality of the structure of a report of an experiment.

Then in your first post you state"

"They ask this on the exam in mass they only had a square box with a full wave with nothing Else in box; but on the bottom of the box which I couldn't read it was too small they had what looked like cm/.011 or something to that defect they ask at the half way point what was the number and then again ask another question asking other questions answers that jumped out at me which they had 4 was 10 and 5 So I picked 5 Didn't do anything like that in training so I don't know Thanks"

This is mostly an incoherent question or statement that I can not really make head or tails of the words and their ordering.

"They ask this on the exam in mass" This I believe is part of an incomplete sentence to tell us you were in Mass. Probably of no importance to the actual question.

"they only had a square box with a full wave with nothing Else in box;" again sort of an incomplete sentence. Probably should have been something like --- A sine wave of X cycles was displayed inside a square outline that might have been the outline of scope face or a plotted graph. No other information was given. Were there no coordinate lines or tick marks associated with the sine wave? Was this a single full cycle starting at a positive slope zero crossing, and ending on a negative slope zero crossing? Or what?

"but on the bottom of the box which I couldn't read it was too small they had what looked like cm/.011 or something to that defect" ---- Was this label inside or outside the outline? What possibly could cm/.011 mean? Probably cm means centimeters. Why cm/.011? More likely I might expect 0.011 sec or volts or something else per cm. But even then it is unusual to have some odd scaling factor per cm. If this labeling is used, then I would expect tick marks or lines on the graph to mark the cm or fractional cm locations.

The rest of your statement seems to be totally incoherent.

Suppose the wave form was a 60 Hz sine wave, then one full cycle is 1/60 second or approximately 0.017 seconds. So that does not correlate with 0.011 . A value of 0.011 would approximately correlate with 90.9 Hz. Thus, possibly they are displaying a 90 Hz wave form, and adjusting the horizontal sweep rate to display the sine wave zero crossings at tick marks.

.

was shown with

- Location
- Ann Arbor, Michigan

To continue.

A typical scope, 5" tube, might be about 10 cm across the screen. So if a full cycle is spread across the full screen, then the frequency is more in the range of 9 or 10 Hz.

.

- Location
- Berkeley, CA

- Occupation
- Retired

- This is a cyclic (repeating) function of period 2. Meaning if you take the graph from 0 to 2 (on the x-axis), and shifted it to the right 2 units, it would perfectly match the graph form 2 to 4. The function from 0 to 2 is called one cycle, from 0 to 1 is called a half-cycle, and from 0 to 1/2 is called a quarter cycle.

- The peak shown is 2 on the y-axis, as labeled. Since the half-cycle is symmetric, the (first) peak occurs half way through the half cycle, or at t=1/2 sec. (t is commonly used for a time variable).

- Now say you drew a straight line between the origin (0,0) and that first peak (1/2, 2). That straight line will lie underneath the graph of the function: at the beginning of the quarter cycle the function is rising faster than its average rate over the quarter cycle; and at the end of the quarter cycle, it is rising slower than average. This can be described as the function being "concave down" over the quarter (or half) cycle.

- It's not clear what exactly your exam question asked. But say it asked (for the above graph) "at what time does the function first hit the value 1, the half-way point to the peak?" If the function were a straight line, it would take half the time to get half way there, so the answer would 1/2 of 1/2, or t = 1/4.

- But we know the function is concave down, and at t = 1/4 the actual function value will therefore be more than halfway to the peak. So the answer to the question is definitely less than 1/4. For a multiple choice question, that might be enough to find the right answer, e.g. if the options were (A) 0 (B) 1/6 (C) 1/4 and (D) 1/3, you know you can pick (B).

- For a sinewave, it turns out the answer is, in fact, 1/3 of the way to the peak, so it is t = 1/6 (for the above graph).

Cheers, Wayne

- Location
- Massachusetts

Excellent explanation, Wayne.

There are several ways to solve for the unknown:

1. Analytically. By constructing the equation of the sine wave, rearranging the terms to make t the subject, and using inverse sine (usually noted as arcsin or sin^{-1} on calculators) to invert the sine function. Pay attention to what angle units your calculator uses (such as degrees or radians), and set up the angular frequency term (the constant B) in the equation below, so that B equals the ratio of a full circle in that angle unit (360 degrees, 2*pi radians, etc) to the measured period.

Steps:

Amplitude A = 2 Volts; Period T = 2 seconds, measured from the graph;

phi = zero phase shift; D = zero DC offset.

Need time t when V = 1 Volt

General equation: V = A*sin(B * t + phi) + D

Constant B = 360 deg/T

Remove the terms equal to zero: V = A*sin(B*t)

Rearrange to make t the subject:

V/A = sin(B*t)

arcsin(V/A) = B*t

t = 1/B * arcsin(V/A)

Recall constant B:

t = T/360deg * arcsin(V/A) in degrees

Evaluate:

t = (2 sec)/360deg * arcsin(0.5 Volts/2 Volts) in degrees

t = 0.16667 sec or 1/6 sec

2. Graphically, by drawing a horizontal line across the graph with a straightedge at the 1-Volt position (the given vertical value for which we want to find the corresponding horizontal value), and locating where it intersects with the waveform. Then draw a vertical line down from this position, to see the corresponding time value, and measure with a ruler to see how the distance from t=0 to this line fits within the given graph scales.

3. Graphically as Wayne described. Identify the key points along the sine wave at each quarter of the cycle. Narrow down the point of interest to where it lies between the key points, and use the shape's slope and curvature for process of elimination to determine which intermediate point is the correct answer.

4. Recognition of special right triangles, as special cases of the trig functions. We're asked for where V = 1/2 of the maximum V, which means we'd like to know when sin(x) = 1/2. This occurs at x=30 degrees, as we know from our 30-60-90 special right triangle. We define a full period to be 360 degrees, so 30 degrees is 1/12th of a cycle. Thus, 2 sec/cycle * 1/12th of a cycle = 1/6 second.

There are several ways to solve for the unknown:

1. Analytically. By constructing the equation of the sine wave, rearranging the terms to make t the subject, and using inverse sine (usually noted as arcsin or sin

Steps:

Amplitude A = 2 Volts; Period T = 2 seconds, measured from the graph;

phi = zero phase shift; D = zero DC offset.

Need time t when V = 1 Volt

General equation: V = A*sin(B * t + phi) + D

Constant B = 360 deg/T

Remove the terms equal to zero: V = A*sin(B*t)

Rearrange to make t the subject:

V/A = sin(B*t)

arcsin(V/A) = B*t

t = 1/B * arcsin(V/A)

Recall constant B:

t = T/360deg * arcsin(V/A) in degrees

Evaluate:

t = (2 sec)/360deg * arcsin(0.5 Volts/2 Volts) in degrees

t = 0.16667 sec or 1/6 sec

2. Graphically, by drawing a horizontal line across the graph with a straightedge at the 1-Volt position (the given vertical value for which we want to find the corresponding horizontal value), and locating where it intersects with the waveform. Then draw a vertical line down from this position, to see the corresponding time value, and measure with a ruler to see how the distance from t=0 to this line fits within the given graph scales.

3. Graphically as Wayne described. Identify the key points along the sine wave at each quarter of the cycle. Narrow down the point of interest to where it lies between the key points, and use the shape's slope and curvature for process of elimination to determine which intermediate point is the correct answer.

4. Recognition of special right triangles, as special cases of the trig functions. We're asked for where V = 1/2 of the maximum V, which means we'd like to know when sin(x) = 1/2. This occurs at x=30 degrees, as we know from our 30-60-90 special right triangle. We define a full period to be 360 degrees, so 30 degrees is 1/12th of a cycle. Thus, 2 sec/cycle * 1/12th of a cycle = 1/6 second.

Last edited:

All impressive but useless for everyday application

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You know of course that if horsepower still came from horses, the plant would have a big opening in the sidewall to bring the hay bales in. You could be grateful everytime you have to carry wire in the door instead of the hay bales. The difference between the two is just an accident in history as to which era you were born into.

I bid a job and wired it per plans and specs, town money town golf course irrigation pumps. The 10 stage 20 hp Goulds pump was on a spec control panel that went to zero flow and held the pump on for three minutes more, shutdown timer, into the closed PRV, the pressure reducing valve.

Let me tell you, that was impressive. That motor would just spin anything connected to it as soon as the 480 was put to it, even the dead headed 10 stage pump pulling pond water and trying to throw it somewhere, even into the closed PRV.

Don't feel bad. I saw the 3 minute timer in the control diagram and tried to get the specialist golf course design engineer to change it, with no useful result. He drove down from out of state to charge them $1000 for the day extra, to tell them to tell me it would work. He was right, it did work. But the PRV needed a rebuild after a few months, the front end was pushed in. Don't know how long the pump lasted but Goulds told me they were pulling the warranty off it when I asked them what would happen to it.

The question on the exam, if they were all like that you would be right that they wanted no one to pass except for the headmaster's nephew and the few others they had sold the answers to in advance. If it's only one or two like that, the question is there to split the 95 percentile guys from the 97 percentile guys.

It would have been solvable algebraically. 1 cycle / second is 1 Hz. 60 cycles / second or 60 Hz is:

60 (cycles) Hz = 1 second

1 (cycle) Hz = 1 / 60 seconds or .0166 seconds

Lots of times you want to know how fast 1/4 cycle is. You want the breaker or fuse to open before peak current. How fast is 1/4 cycle at 60 HZ ?

1/4 (cycle) Hz = (1 / 60) / 4 seconds or .00416 seconds, convert that to milliseconds is 4.16 milliseconds.

That's about 1st year HS algebra along with some dimensional analysis. Also, the question is there to see if you get frustrated by it (then need a cigarette), or if your adrenaline is up and you're not going to let a question like that kick your butt. lf you find learing something exciting, or if you like constantly rebuilding the same PRV.

Also, the teacher's need someone to carry the hay bales for them, so I'm sure they get some of the blame also.

- Location
- Ann Arbor, Michigan

It would be interesting to see an actual photocopy of the question as presented to Scottywatt.

As an alternative is there anyone else that has encountered this question that could fill in the details?

The question is unanswerable or there are details of which we are unaware.

.

- Location
- Anchorage, AK

- Occupation
- Engineer

This may help in finding time of a sine wave.

f = 1 / T ;

f = frequency

T = time of cycle (period)

f = 1 / T ;

f = frequency

T = time of cycle (period)