This is a little, possibly obvious lesson that took me a surprisingly long time to learn.

It's one of those things that for me was obvious in retrospect, but just took me ages to get there.

Context: It was while playing around with the rocketry equation u/v = exp(w/z)-1 for Ken that I bumped my head against this issue.

(Let's also introduce the symbol R = exp(w/z) = u/v +1 .)

(it doesn't matter what the variables are for the point of discussion, though I can explain it if necessary)

Anyway, while it doesn't really solve that problem directly, imagine that I am trying to split this into two nomograms with a common scale:

g(U,V) = t and t = h(W,Z)

In this case it would be nice to be able to set ln(u) + log(v) = ln(R-1) on one side and take logs, and then w/z = ln(R) on the other, but this can't be done - at least not without one of those ugly scale-change ladders in between the two R-scales:

|/|

|/|

| |

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|\|

But lets imagine I manage to transform both sides of the equation (at least to a high degree of approximation) so that I have

g(U/V) = t and t+c = h(W/Z), where in each case, t is some function of R.

Well, the c turns out not to be a problem - of course, you can take c over the other side, though sometimes this is an inconvenience.

However, it can be left in this form - just construct the two nomograms as is.

Imagine you had a pair of scales for t, the left one with t on it and the right with t+c. You could imagine a scale-ladder connecting them.

|\|

|\|

|\|

|\|

(c might be positive or negative, so this scale might be going either way)

Now move the left or right nomogram up or down so that its rungs are perfectly horizontal:

|-|

|-|

|-|

|-|

|-|

and smoosh them together into a single scale. Maybe it's obvious. But it took me a while. Shifting one side of the nomogram up or down to make them line up simply corresponds to taking the constant over the other side of the equation - so we're just doing what we could have done, but looking at it a different way.

We don't actually need this "t" (/ "t+c" ) scale to be marked. However, maybe we want the to mark the t scale (after all, it's a function of R on either side, and R is meaningful in the original problem, even if it's not our primary interest). The thing is, we can mark the R values or the R-1 values - or both - on their appropriate side of the scale.

[Once you're done, you have to double check - use the nomogram for calculations and see that they give the right answers, because it's easy to make a mistake and stuff something up if you're moving scales about without actually going back to the original mathematics and doing it all. In fact, this is a good rule with any nomogram - try a number of test cases, because it's easy to stuff up).

It is possible to do all of this for the rocket equation - produce a pretty good approximate nomogram for the left and right halves of this equation with a common scale in the middle (at least for the range of values I was playing with when I was thinking about this problem).

But it led me along a bit further. What if I wanted to make a nomogram for which (as an example) the left half was like so:

|/| <- t

and for which the other half was like

t -> /|\

Again, it's possible to draw them as is, and then just rotate the two diagrams to bring the t-scales into alignment, like so:

...t__

|/|\

Mathematically, it is straightforward to apply this rotationto the determinant to do the same thing as to physically rotate one of the two diagrams, but it wouldn't have occured to me to do this.

See, some books say something along the lines of "the common scale has to be the same determinant-row in both determinants" which is true, but that sort of statement was "blocking" my ability to see that it's often easy to make it so. I would find myself focusing on a single t-row and trying to write both equations to fit with some single one.

That's too hard! (at least, it was for me)

It's easier to start with two separate nomogram equations, and while keeping in mind that the transformation of the variable has to end up the same, you can play around with them separately. If the scale for each is linear, or logarithmic or whatever, you may be able to make them fit together, even though the current parameterization doesn't quite fit.

The fact is that any of the matrix transformations may be applied to either determinant in order to eventually make the shared scale the same. Even though mathematically it's no different, I found being able to play with the two and then find a way to transform them to share a scale substantially easier to make progress with.

That is, affine transformations - rotations, scaling, translation, reflection, skewing and so on are all fair game - but also projections can be used. If one t-scale is parameterized by the row

[0, t, 1] and the other t-scale by the row

[at/(t+1), b - kt/(t+1), 1]

they are not the same parameterization. My old way of thinking about shared scales would have immediately rejected this possibility, but of course they can be transformed one into the other.

So I guess the short lesson is you don't have to lock yourself into starting with a common scale for the shared variable. You can explore the nomogram possibilities a bit more, if you keep in mind as you go that as long as there's a way to transform the two sets of scales to be the same, you can play about somewhat more freely with them.

## Shared scales

### Anamorphosis with back to back scales

The usual double-anamorphosis (due to Lafay) that is in the books is generally better than the simpler single anamorphosis - which can still be useful when all of the curves on the intersection chart are similarly curved.

However, there is a situation in which adjusting a single scale is more useful than the more powerful form - when you have compound charts with a shared scale. In that case, you may not be able to change the common scale; adjusting the other straight scale can then solve the problem.

Lyle (see the thread on his paper in the Literature subforum) describes in detail how to do this simpler form of empirical adjustment first.

This was the approach used for the annuity-value (mortgage) nomogram I discussed in another post - I tried a square root scale for i and a log scale for a, which made the n-curves in the intersection chart almost straight, apart from at the lowest interest rates. A small adjustment of the locations of the values of the lowest interest rates made the curves almost all as straight as I could discern (this was easy in Excel - I could push them around by literally changing their value until things were nice). Only the lowest few values required adjustment. Then with the n-scale all close to straight lines, the two straight scales were plotted in parallel, and the curved scale fitted by drawing three lines (at high, low and medium interest rate for a given value of n and the corresponding values of a) and marking a cross in the little triangles that resulted.

However, there is a situation in which adjusting a single scale is more useful than the more powerful form - when you have compound charts with a shared scale. In that case, you may not be able to change the common scale; adjusting the other straight scale can then solve the problem.

Lyle (see the thread on his paper in the Literature subforum) describes in detail how to do this simpler form of empirical adjustment first.

This was the approach used for the annuity-value (mortgage) nomogram I discussed in another post - I tried a square root scale for i and a log scale for a, which made the n-curves in the intersection chart almost straight, apart from at the lowest interest rates. A small adjustment of the locations of the values of the lowest interest rates made the curves almost all as straight as I could discern (this was easy in Excel - I could push them around by literally changing their value until things were nice). Only the lowest few values required adjustment. Then with the n-scale all close to straight lines, the two straight scales were plotted in parallel, and the curved scale fitted by drawing three lines (at high, low and medium interest rate for a given value of n and the corresponding values of a) and marking a cross in the little triangles that resulted.

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