I would say the cure is worse than the disease. It is easier to do the real thing. Remember, the purpose of the approximation is only to check your results. If you did it the wrong way it should be obvious.

 

I can never remember if its product over sum or other way.
but I do remember two identical resistors in parrale is half the resistance, so plug in 1 for r1 and r2, and check !

Again, just track the units.

Product over sum means you have Ω² in the numerator and Ω in the denominator, leaving you with Ω, which is what you want.

Sum over product means you have Ω in numerator and Ω² in the denominator, leaving you with 1/Ω, which is NOT a unit of resistance.

 

Again, just track the units.

Product over sum means you have Ω² in the numerator and Ω in the denominator, leaving you with Ω, which is what you want.

Sum over product means you have Ω in numerator and Ω² in the denominator, leaving you with 1/Ω, which is NOT a unit of resistance.

Was just saying how I remembered / checked it.

I've heard many other ways as well

 

I would say the cure is worse than the disease. It is easier to do the real thing. Remember, the purpose of the approximation is only to check your results. If you did it the wrong way it should be obvious.

Approximations are also used to do estimates without calculating the results first.

On-the-fly situations require you to understand the concepts well enough that you can come up with a decent estimate without relying on memorized estimation formulas.

Situations in which you routinely need to do quick estimates of the same kind can benefit from spending time up front to figure out, and then memorize, an acceptably accurate estimation algorithm. As a pilot, there are a number of scenarios where this is at play, such as doing wind corrections, fuel calculations, time estimates for various maneuvers. You could pull out a calculator (or the old E6B), but often don't have the time and have too many other things going on at the same time. Long range rifle marksmanship is another example.

As with all things in engineering, the estimation algorithm must be as complex as needed to yield results that are accurate enough, but should generally not be any more complex than that.

 

Something to keep in mind is not making the cure worse than the disease.

Let's compare

Rp ≈ sqrt(R1*R2)/2

and

Rp = (R1*R2)/(R1+R2)

In both cases, you have to multiply R1 and R2 together. In both cases you need to divide one thing by something else, though in the approximation you are dividing by two, which is something that is easily done on paper. The only other differences is in the approximation you have to take a square root, not something that most people can do accurately or quickly by hand, while in the actual formula you merely have to add two values together. So while evaluating the actual result is quite doable, even by hand, for most people, evaluating the approximation is something that most people wouldn't do unless they have a calculator hand and, if they do, why would the settle for an approximation instead of just finding the actual value.

In the case that one resistor is within a factor of two of the other, we already know that

Rmin/2 ≤ Rp ≤ Rmax/2

Where Rmin is the smaller of the two and Rmax is the larger.

What if we just use the arithmetic mean as our approximation:

Rp ≈ [(R1+R2)/2]/2 = (R1+R2)/4

Now you have something that most people should be able to do readily on paper and be able to get a close answer in their head. The error using the arithmetic mean over this range is almost exactly twice the error of using the geometric mean, peaking at 12.5% when one resistor is twice the other, compared to 6.1% for the geometric case.

An approximation is only useful if it involves sufficiently less effort to achieve than doing the full-blown evaluation, so while using the geometric mean might result in a closer approximation, why would anyone ever actually use it?

If we want a better approximation, while keeping in mind the need to keep it reasonably simple to calculate, we can leverage the known behavior of the error. In this case, if the larger resistor is half-again as large as the smaller, we have an error of just over 4% at that point, growing to just over 12% when it is twice. So, what if we simply reduce the estimate by 5% (i.e., one-half of 10%) when we cross that threshold?

Example: R1 = 470 Ω, R2 = 750 Ω.

Without the adjustment,

Rp ≈ (470 Ω + 750 Ω) / 4 = 1220 Ω / 4 = 305 Ω

The difference between them is 280 Ω, which is more than half of 470 Ω, so we have to apply our adjustment.

10% of 305 Ω is 30.5 Ω and half of that is 15.25 Ω, so our adjusted estimate is

Rp ≈ 305 Ω - 15.25 Ω = 289.75 Ω

The actual value is 288.93 Ω, so our estimate, which was arrived at without using a calculator or even writing anything down beyond what I typed above, is off by 0.28%.

Using the geometric mean estimate

Rp ≈ sqrt(470 Ω · 750 Ω) / 2

I can simplify this some by pulling a factor of 100 out of the sqrt

Rp ≈ 5 · sqrt(47 Ω · 75 Ω)

While multiplying 47 by 75 is not terribly difficult to do on paper, it's not something that I'm going to do in my head. If I had to, I would probably estimate it as roughly 50*70 = 3500, which would allow me to pull another factor of 100 out, giving me

Rp ≈ 50 · sqrt(35) Ω

Now, I know that sqrt(35) is very close to 6 and slightly less, but I don't have a good feel for how much, particularly in my head, so I might call it 5.9, or a reduction of 1/6 of 10%. With that, I would get

Rp ≈ 300 Ω - 30/6 Ω = 295 Ω

A lot of work, if I don't have a calculator, to get an estimate that is 2.1% off.

If I have a calculator, I get an estimate of 296.86 Ω, which is actually worse (but that's by coincidence) and is off by 2.7%, which is right at an order of magnitude off from the estimate obtained using the arithmetic mean, with adjustment, that was obtained doing all the math in my head. Note that, without adjustment, the arithmetic mean estimate of 305 Ω is, as expected, about twice as far off as the geometric mean with an error of 5.7%.

Remembering the rule of reducing the arithmetic mean estimate by 5% if the larger resistor is more than 1.5x the smaller is easy (and easy to do), it still results in a max error of about 6.8% (comparable to the geometric mean estimate). It isn't too difficult to improve this by adding more break points to the adjustment. How many and were would be a balance between computability and accuracy, always keeping in mind that the computability component is the overriding factor in order to avoid the cure becoming worse than the disease.

Hi there,

Well then I guess for better or worse you did actually find that an interesting method

I agree with all or at least most of what you said and liked that post, but I was actually under the impression that we were looking for methods, not necessarily approximations for the sake of approximations. This led me to explore ideas that may even be harder than the original as long as they were in fact methods, as long as they were not so trivial like squaring everything and then taking the square root just to claim it's another method (although that results in the absolute value in a program without an if-then structure).

But if you didn't like that method (it's ok though no problem) you must of missed #3 in post 13 which you would literally hate

And just to be clear, whether you agree or disagree I always like reading your posts, especially when they are as comprehensive as that one was.