# I am a math rock. Simple transposition question

Discussion in 'Math' started by SgtWookie, Feb 27, 2010.

1. ### SgtWookie Thread Starter Expert

Jul 17, 2007
22,182
1,728
Terrible to admit, but I'm a math rock.

I've been struggling with this for a day, and I simply don't "get it" - this is high-school stuff, and that was a long time ago. Where am I going wrong?
The idea is to get L over on the left side of the equation.

Somehow I've managed to be involved in complex electronics and software for many years, but frankly I suck at math. I don't understand where I'm going wrong here.

Fo=1/2pi*sqrt(Lx*Cs) (this works)
L =4*pi()^2*((1/(D73*E73/1E6))^2/5e-8) (wrong!)

Various things I've tried:

To=2pi*sqrt(Lx*Cs)
To*2pi^2=4pi^2*Lx*Cs

=((d73*e73/1e6)/pi()*2)^2/5e-8

To=2*PI()*SQRT(F69*0.05) <=that works; F69=L(uH)
0=2*PI()*SQRT(F69*0.05)/To
0/2*PI()^2=(F69*0.05)^2/To
0/2*PI()^2/F69^2=0.05^2/To

To=2*PI()*SQRT(F69*0.05) <=that works; F69=L(uH)
To=2*PI()*SQRT(F69*0.05/1E12) <=that works
F=1/(2*pi()*sqrt(F66*0.05/1E12)) <=that works
F=1/(2*pi()*sqrt(F65/1E6*0.05/1E6)) <=that works

L=4*pi()^2*(L67^2/(.05/1e6)) <- garbage
L=4*pi()^2*((1/d68*e68)^2/.05) <- garbage
Where am I going wrong?

2. ### hgmjr Moderator

Jan 28, 2005
9,030
214
$f_o=\frac{1}{2*\pi*\sqrt{(L*C)}}$

Square both sides:

${f_o}^2=\frac{1^2}{{4*\pi^2*{\sqrt{(L*C)}}^2}$

Simplify

${f_o}^2=\frac{1}{{{(4*\pi^2*L*C)}}}$

Multiply both sides by L:

${f_o}^2*L=L*\frac{1}{{{(4*\pi^2*L*C)}}}$

L cancels on the right:

${f_o}^2*L=\frac{1}{{{(4*\pi^2*C)}}}$

Divide both sides by ${f_o}^2$

$L=\frac{1}{{{(4*\pi^2*C)*{f_o}^2}}}$

Substitution allows us to exhange omega for 2*pi*f

$L=\frac{1}{{{(\omega^2*C)}}}$

We are all rocks in the math quarry. It is just a matter of size.

hgmjr

Last edited: Feb 27, 2010
3. ### SgtWookie Thread Starter Expert

Jul 17, 2007
22,182
1,728
Thanks for your reply hgmjr, which I'm sure took a lot of work to get all of the tex graphcs right. But I still don't know how to get L over to the left side, but more importantly, how to do so.

Like I said, I'm a math rock. You've tried to provide instructions that would be adequate for most, but for some reason I have a really hard time with it.

Thanks,
Wook

4. ### hgmjr Moderator

Jan 28, 2005
9,030
214
Which of the steps is not clear? Maybe I can break it down a bit more.

hgmjr

5. ### someonesdad Senior Member

Jul 7, 2009
1,585
141
The key idea in manipulating equations is that you have to do the same thing to both sides -- otherwise it doesn't remain an equality.

Thus, suppose you have an equation

$a = \frac {b} {L}$

and you want the L on the other side. The strategy used is to perform the same operation on both sides so that the L disappears from the right side and goes to the left side. Here, we can multiply both sides by L:

$L a = \frac {b} {L} L$

which reveals our strategy, because L divided by L is just 1. To see this, rewrite things slightly (using the associative law of multiplication) to get

$L a = b \frac {L} {L}$

Thus, we "cancel out" the L's on the right and we are left with the equation

$L a = b$

Analogously, to move the a to the right hand side, we just divide both sides by a to get

$L = \frac {b} {a}$

6. ### hgmjr Moderator

Jan 28, 2005
9,030
214
Someonesdad has captured the essence of the technique needed to manupulate the equation indeed any equation. This particular equation has an added factor in that L is contained in a square root term. This is what required the squaring of both sides of the equation to liberate L from the square root term. This squaring of both sides is consistent with someonesdad's statement that what is done to one side of the equation must be done to the other side of the equation if the equality is to be maintained.

hgmjr

7. ### Louise New Member

Jan 17, 2010
13
1
If I may be so bold as to make a tiny suggestion - something that helps me sometimes (and my maths ability might be accurately described as 'vanishingly small') is to try to re-arrange the same equation using numbers instead of symbols. The same rules apply.

8. ### Wendy Moderator

Mar 24, 2008
20,735
2,499
I figure we are as close as two people can be technically with such different back grounds. I find I'm having trouble with a lot of the math at this site. I do what can, and ask for help with the rest, same as you.

There was a time I would claim I knew some calculus. However, it rusts like the rest of my skills, but I really don't have a use for it very often. It is pretty much gone now.

9. ### hgmjr Moderator

Jan 28, 2005
9,030
214
If it helps. The above attempt came off the tracks when both sides of the equation were divided by To in STEP 2. The left side of the equation should have been 1 rather than 0. This is because To/To = 1.

I think it is safe to assume that in these equations references such as F69 and F66 are artifacts of the cut and paste operation from the spreadsheet you are using.

Everything you did above was kosher.
In the above box, I can't easily interpret what the cell references equate to so I am unable to determine where things went south.

Hope this helps some.

hgmjr

10. ### SgtWookie Thread Starter Expert

Jul 17, 2007
22,182
1,728
Thanks folks, I figured it out.

It's just a simple Excel spreadsheet that I've been using for testing an assortment of toroids I've had kicking around here for awhile.

I'm using a 74HC14 hex Schmitt-trigger inverter IC with three 100nF caps and a couple of resistors as a buffered Pierce oscillator; from an article published in EDN's Design Ideas, and measured the time of the output square wave with my O-scope (haven't built a freq. counter yet; still in the works...).
EDN article: http://www.edn.com/article/CA6462564.html

I record the divisions and time/div, and out pops the inductance in uH and approximate AL value. It's pretty decent down to maybe 10uH, but then the propagation delay in the 74HC14 starts kicking in along with parasitics from the X7R caps I used. A known 977nH inductor measures 2uH, so it's not all that bad.

One of the caps is used as a supply bypass cap for the 74HC14. The other two caps are on either side of the inductor under test, so they're essentially in series for a total of 50nF.

I was forgetting to shift my decimal places around where they belonged. Adding in a few /1e6 and *1e-6 fixed things up.

L(uH)=1/(4*PI()^2*(1/(scope_divisions*uSecs_per_division*0.000001))^2*(0.05*0.000001))*1000000

11. ### Harrington New Member

Dec 19, 2009
86
3
Here try this I'm not brilliant at maths myself but this the way we solve problems like this see attached hope this helps you in the future with your calculations

I use this from time to time you can also solve matrices maths with this and calculus Its a dos version but it will run on windows XP

Hope this helps you

Mark

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