# insights into all 70% marks i.e. rms, 3dbrolloff, ect

#### zachary schexnaydre

Joined Apr 18, 2013
4

Has anyone ever heard of this concept?
Is there a name for this concept?
Also is this taught in text books? Am i over analyzing this?

#### Markd77

Joined Sep 7, 2009
2,806
You might be on to something.
The real answer is peak = (square root of 2) X RMS
so RMS = 0.707107..... X peak
I just did a quick experiment in a spreadsheet with numbers 1-1000 and it turns out that the closest I got was sum(1:707) is approximately sum(708:1000)
You may have found a new way of calculating 1/sqrt(2), although it's possible that it isn't quite right or that it has been discovered before.

#### t_n_k

Joined Mar 6, 2009
5,455
If you equate the two integrals

$$\int_0^{kX}{x.dx}=\int_{kX}^X{x.dx}$$

you find

$$k=\frac{1}{\sqrt{2}}=0.707$$

#### Markd77

Joined Sep 7, 2009
2,806
Nice, unfortunately I forgot how to do integrals years ago.

#### Papabravo

Joined Feb 24, 2006
16,486
Don't worry about the integrals if you're not familiar with them.
RMS is an acronym for Root Mean Square

The way to think of this is
1. Divide a cycle of the waveform into equal time intervals
2. Square the value at each interval
3. Sum up all of the squared values
4. Divide the sum of the squares by the number of intervals
5. Take the square root
This is a way of getting a non-zero average that just happens to give the useful result that RMS votage times RMS current gives average power. So .707 is the square root of the average (aka mean) of the sum of the squared values. Any other relationship that you see is just a coincidence and means practically nothing.

#### zachary schexnaydre

Joined Apr 18, 2013
4
so I added 1 to 700 and got 244,576 and then I did 701 to 1000 and I got 245,215 which is only 639 apart which is also only.26% of 245,215 so at 70% it seems extremely close but not so much at the 70.7% like you tried. so I'm thinking that this might only work at 70%. Any feedback to this would be highly appreciated.

#### GopherT

Joined Nov 23, 2012
8,012
If you equate the two integrals

$$\int_0^{kX}{x.dx}=\int_{kX}^X{x.dx}$$

you find

$$k=\frac{1}{\sqrt{2}}=0.707$$
Interestingly, a piece of paper with length and width $$1:\sqrt{2}$$ will always have the same proportions after it is folded in half (and rotated 90 degrees).

#### Papabravo

Joined Feb 24, 2006
16,486
so I added 1 to 700 and got 244,576 and then I did 701 to 1000 and I got 245,215 which is only 639 apart which is also only.26% of 245,215 so at 70% it seems extremely close but not so much at the 70.7% like you tried. so I'm thinking that this might only work at 70%. Any feedback to this would be highly appreciated.
Remember that 0.707... is just and approximation to three decimal places of an irrational number. 70% is an even worse approximation.

#### Markd77

Joined Sep 7, 2009
2,806
so I added 1 to 700 and got 244,576 and then I did 701 to 1000 and I got 245,215 which is only 639 apart which is also only.26% of 245,215 so at 70% it seems extremely close but not so much at the 70.7% like you tried. so I'm thinking that this might only work at 70%. Any feedback to this would be highly appreciated.
sum 1-707 = 250278
sum 708-1000 = 250222
but the integral above shows that it gets closer to 1/(sqrt(2)) or about 0.707 as the numbers get bigger.
It seems amazing luck that you came up with this thinking that the correct answer was the approximation of 70%, but somehow got the exact correct figure.

#### Papabravo

Joined Feb 24, 2006
16,486
If you equate the two integrals

$$\int_0^{kX}{x.dx}=\int_{kX}^X{x.dx}$$

you find

$$k=\frac{1}{\sqrt{2}}=0.707$$
You should really ask: "What is the integral of sin squared (x)dx from 0 to 2π divided by 2π. The answer is 1/2 and the square root of that is... you guessed it 0.707 or 1 over √2.

#### zachary schexnaydre

Joined Apr 18, 2013
4
Ok now not referring to rms, just in general why is it that the upper 30% is always really close to the bottom 70%? Is there some mathematical formula showing this or do u really think that this is coincidental? also if this something to do with an already existing math formula can you tell me the name of it?

#### zachary schexnaydre

Joined Apr 18, 2013
4
thanks I think you are right about it getting more accurate with the .707. I also think that its a really cool thing how it works out this way.

#### Papabravo

Joined Feb 24, 2006
16,486
Maybe you can make some generalization around Euler's (pronounced Oiler) formula for the sum of an arithmetic series.

The sum of n integers is ((n)(n+1))/2
sum(1:10) = ((10)(11))/2 = 55
sum(1:7) = ((7)(8))/2 = 28
Therefore the sum(8:10) must be 55 -28 = 27
Is it coincidence or magic? You tell me.

#### t_n_k

Joined Mar 6, 2009
5,455
@zachary,

The process of adding a very large number of integers using the rules as you have shown begins to approximate the addition of any range of real numbers using the same rules of comparison.
While this is an interesting mathematical curiosity I believe it has little relevance to RMS values of sinusoidal functions.

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#### t_n_k

Joined Mar 6, 2009
5,455
You should really ask: "What is the integral of sin squared (x)dx from 0 to 2π divided by 2π. The answer is 1/2 and the square root of that is... you guessed it 0.707 or 1 over √2.
I think you may have missed the point of the relationship. This [as in my first post] merely shows the condition where the summation of real values over some fraction k of a real value range [0,X] will equate to the summation of real values in the range of [0,X] not including 0 to kX {i.e. the summation of [kX,X]}. The relevant value of k which satisfies the equality just happens to be 1/√2 when summation [0,kX] equals summation [kX,X]. I wasn't attempting to derive the RMS value of a sinusoid. I was hoping rather to describe succinctly what the OP had proposed as something of interest in relation to the summation of integers under particular constraints. I suggest that the extension of the concept from the set of integers to the set of real numbers leads to the definitive relationship.

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#### YokoTsuno

Joined Jan 1, 2013
43
Hi Zachary,

Ok now not referring to rms, just in general why is it that the upper 30% is always really close to the bottom 70%? Is there some mathematical formula showing this or do u really think that this is coincidental? also if this something to do with an already existing math formula can you tell me the name of it?
The 70.7% is coincidental. The only thing these figures have in common is that both originate from a square somewhere in the calculation (1/√2).

Your summation creates the impression that this is some sort of series expansion. In reality you actually perform a discrete integration, in this particular case the summation of all rectangles of width 1 on a triangle (area below y = ax).

Like this: Total = (1x1) + (2x1) + ........... (700x1).

The fact that it is discrete (rectangles are not infinitely small) is also the reason why you get the 1-3% error.
Mathematically you actually calculate a specific point (point z, @ 70.7%) where the area above z equals the one below z.

y = ax
ʃax = ax²/2

Hence:

ax²/2 over [0-z] = ax²/2 over [z-100]
az²/2 - 0 = (a/2).100² - (a/2)z²
z² = 100² - z²
2z² = 100²
z = 100/√2 = 70.7 generally 70.7% or 0.707

The "trick" you apply would not work anymore for functions other than y = ax since the √2 implies linearity.

b) RMS value:
This √2 (70.7%) has a very different origin. The RMS value of an AC voltage is the value of an equivalent DC voltage which creates the same amount of heat in a resistive load. P = U²/R. Since the square of a sine wave is:
- Always positive
- Entirely symmetrical
See here: http://en.wikibooks.org/wiki/Trigonometry/Graph_of_Sine_Squared

(Just shift the x-axis to the amplitude of the sin²x)

.. the power generated is therefore only 50% of the peak of the squared sine and the corresponding RMS voltage obviously the square root of this 50% or 70.7%.

I am a actually bit surprised that your teachers fail to explain this.

#### Papabravo

Joined Feb 24, 2006
16,486
I think you may have missed the point of the relationship. This [as in my first post] merely shows the condition where the summation of real values over some fraction k of a real value range [0,X] will equate to the summation of real values in the range of [0,X] not including 0 to kX {i.e. the summation of [kX,X]}. The relevant value of k which satisfies the equality just happens to be 1/√2 when summation [0,kX] equals summation [kX,X]. I wasn't attempting to derive the RMS value of a sinusoid. I was hoping rather to describe succinctly what the OP had proposed as something of interest in relation to the summation of integers under particular constraints. I suggest that the extension of the concept from the set of integers to the set of real numbers leads to the definitive relationship.
I did not miss the point at all, but in fact was unaware of that particular relationship. I don't think it has any relevance to the issue of why we use the RMS calculation for the "average" value of certain periodic waveforms. It is an interesting relationship especially if you substitute other functions for x and dx. For example if x = sinθ and dx=cosθdθ your relationship is still true.

#### t_n_k

Joined Mar 6, 2009
5,455
I did not miss the point at all, but in fact was unaware of that particular relationship. I don't think it has any relevance to the issue of why we use the RMS calculation for the "average" value of certain periodic waveforms. It is an interesting relationship especially if you substitute other functions for x and dx. For example if x = sinθ and dx=cosθdθ your relationship is still true.
Thanks for the feedback Papabravo. It's probably not worth pursuing this further, as I guess most of us who have commented don't believe there is a link between the OP's proposed "conundrum" on the manipulation of integers and the RMS value of sinusoidal functions.

I'm not sure that if one applies the function substitution you suggest for x and dx that the relationship I posted yields the same value for the unknown factor k - for all values of limit X [or its equivalent angular value with the change of variable]. In any event, it seems to me this topic has no special interest for the Electronics Chat forum.