Translated from Rivista Tecnica di Aeronautica, N. 3, May—June 1911, By F.W. MORRISON and C.W. FREDERICK.

1. The altitude of an aeroplane may be determined in the following manner:*

Establish a horizontal base AB (Fig. 1), and let P be the aeroplane, and H its projection on the horizontal plane passing through AB. With two theodolites measure the angles of elevation

HAP=a', HBP=?’,

and the angles in azimuth

HAB=a, HBA= ?.

Then, assuming AB as the unit of length, let

HP/AB = h AH/AB=m BH/AB=n

* Henri Petit: Comment on mesure l’altitude atteinte par les aeroplanes. "La Nature," 19 Novembre, 1910.

m=sin ?/sin(a+ ?) (1)

n=sin a/sin(a+ ?) (2)

h=m tan a’ (3)

h=n tan ?’ (4)

whence

h=sin ? tan a’/(sin a+ ?) (5)

or

h=sin a tan ?’/sin (a+ ?) (6)

We propose to construct a nomogram which will do away with the numerical computation of the above formula. Among the different types that might serve for such a purpose, we believe the most suitable is that obtained by superposing two nomograms with circular scales.* The principle of their construction is so simple that it can be understood even by those who are unacquainted with the elements of nomography. Thus we hope to increase to some extent the interest in this new and fertile science.

NOTE.—It would evidently be sufficient to know only one of the two angles a' or ?', but it is well to know them both, either to serve as a check up on the two values of h which can be obtained in this way, or for cases where the sine of one of the two angles a' or ?' may be in determinate.

2. In the equation

t_{1}f_{2}+f_{3}=o (7)

let f_{1} represent any function of a single independent variable z_{1}, and similarly let f_{2} be any function of z_{2}, and f_{3} and z_{3}. The equation can be written in the form of a determinate as follows:

1 f_{1} 1+f_{1}^{2}

1 f_{2} 1+f_{2}^{2} =o (7’)

1 o 1+f_{3}

* Nomograms of superposed conical scales were brought to notice by Clark, Professor at the Politecnico di Cairo. See Theorie generale des abaques d'alignement. "Revue de Mecanique," 1907. Also see D'Ocagne, Calcule graphique et Nomographie. "Encyclopedie scientifique", Doin.

as may be easily verified. Then referring to a system of rectangular axes,* the three points

x_{1}=1/1+f_{1} y_{1}=f_{1}/1+f_{1}^{2} (8)

x_{2}=1/1+f_{2} y_{2}=f_{2}/1+f_{2}^{2} (9)

x_{3}=1/1+f_{3} y_{3}=o (10)

are seen to be in a straight line. From (8) it is clear that upon varying z_{1} the locus of the points (x_{1}, y_{1}) is a circle, for by eliminating f_{1} we obtain

x^{2} + y^{2} - x=o

From (9), upon varying z_{2}, the locus of the points (x_{2}, y_{2}) is the same circle. And from (10), varying z_{3} the locus of the points (x_{3}, y_{3}) is the axis of abscissas.

Having shown this, let us give to z_{1} a series of values (in arithmetical progression, for example), and by means of equations (8) compute the coordinates (x_{1}, y_{1}) corresponding to each value of z_{1}. When we plot these points they will fall on the circle mentioned above, and we may assign to each the corresponding value of z_{1}, which will then be its scale value. Thus we obtain scale (z_{1}). Similarly if we vary z_{2}, we shall obtain scale (z_{1}), upon the same circle. And by varying z_{3}, we shall obtain scale (z_{1}), upon the axis of abscissas. These three scales, when plotted, constitute a nomogram.

It is now evident, that, given two of the three quantities z_{1}, z_{2}, z_{3}, which satisfy equation (7), as z_{1} and z_{3}, in order to obtain the third, we determine the point of scale (z_{1}) which represents the scale reading z1, and the point of scale (z_{2}) which represents the reading z3. The straight line joining these two points will intersect the scale (z_{2}) at a point the scale reading of which will be the required value of z_{2}.

The reading of the scales will generally require an interpolation by eye.

NOTE I.—The accuracy attained will always depend upon the dimensions of the figure. -

NOTE II.—To construct the scales (z) we could also proceed graphically,? but for the sake of accuracy, and the checking of

* D'Ocagne, see p. 289.

? D'Ocagne, see pp. 171 and 290.

the figure, it is best to compute all three of the scales numerically from the equations (8), (9), and (10).

NOTE III.—Instead of tracing the connecting line upon the design, it is better to use a fine thread.

3. With the above as a basis, we see that by introducing two arbitrary constants k_{1} and k_{2} we may write equation (1) in the form

k_{1}m k_{2}/k_{1} sin (a + ?) –k_{2} sin ? =o

which is similar to equation (7), if we make

z_{1}=m f_{1}=+k_{1}m

z_{2}=a+? f_{2}=- k_{2}/k_{1} sin (a+?)

z_{3}=? f=+k_{2} sin?

With these [or similar equations from (2)] we can construct a nomogram as explained in the previous paragraph.

Introducing another arbitrary constant 1, the scales of this nomogram may be computed by writing equations (8), (9) and (10) in the following form:

X_{1}=+ l/1+(k_{1}m)^{ 2} y_{1}=+ lk_{1}m/1+(k_{1}m)^{ 2} (11)

X_{2}=+ l/1+(k_{2}/k_{1})^{ 2} sin^{2} (a+?) y_{2}=- lk_{2}/k_{1} sin (a+?)/ 1+(k_{2}/k_{1})^{ 2} sin^{2} (a+?) (12)

X_{3}=+ l/1+k_{2} sin ? y_{3}=o (13)

The common basis of the two scales (m) and (a+?) will be a circle with center at (l/2, o), and radius l/2.

Thus we obtain Fig. 2, in which we have taken l=4 cm. and k_{1}=k_{2}=1. From this nomogram, if we put a=130°, and ?=300, we find m=1.4.

NOTE 1.—The values of y_{1} and y_{2} can be obtained from those of x_{1} and x_{2} by noting that

y_{1}=[k_{1}m]x_{1}, y_{2}=-[k_{2}/k_{1} sin (a+?)]x_{2},

and the computation of x_{2} can be simplified through the well- known relation

sin^{2}(a+?)= ½ [1=cos^{2} (a+?)]

NOTE II.—Having computed scale (?), the scale (a+?) may be obtained from it, for when m=1 we have from equation (1) sin/sin(a+?), whence reducing (a+?) to the first quadrant (if necessary), the points of the two scales having the same value must fall on a straight line with the point m=1. For the sake of accuracy, however, it is advisable to make use of this property only as a check upon the computations.

4. Introducing another arbitrary constant k_{3}, the equation (3) can be put in the form

k_{1}m k_{3}/k_{1} 1/h – k_{3} cot a’=o

which is similar to equation (7), if we put

z_{1}=m f_{1}=+k_{1}m

z_{2}=h f_{2}=- k_{3}/k_{1} 1/h

z_{3}=a’ f_{3}=+k_{3} cot a’

With the above [or similar equations from (4)] we can construct a nomogram like the preceding one. If we give to l and k_{1} the same values they had in the preceding case, the new scale (m) will be the same as before.

Scales (h) and (a') are then obtained from the formulae

x_{4}=+ l/1+(k_{3}/k_{1})^{ 2} 1/h^{2 } y_{4}=- l k_{3}/k_{1} 1/h / 1+(k_{3}/k_{1})^{ 2} 1/h^{2} (14)

x_{5}=+ l/1+k_{3} cot a’ y_{5}=o (15)

Thus we construct Fig. 3, in which l and k_{1} have the same value a sin the preceding figure, and k_{3}=1. From this nomogram, if we put m=1.4 and a'= 50°, we find h=1.6.

NOTE II.—The values of y may be obtained by noting that

y_{4}=-[k_{3}/k_{1} 1/h]x_{4}

Also, having constructed the scale (a') from o° to 45°, we see that when k_{3}=1 the remaining part of the scale is symmetrical with the first in respect to the center of the circle, for it is evident

l/1+cot a’ + l/1+cot(90°-a’) = l

NOTE II.—If it were not for the checking mentioned before (Art. 2, Note II, and Art. 3, Note II), having computed scale (m), the calculation of scale (h) would be unnecessary, for two reasons: First, when a'=.45° we have h=m; consequently the points which in the two scales have equal values, are in a straight line with the point a'=45°. Second (true only when k3=1), for equal values of h and m we have

x_{1}+x_{4}=l and y_{1}=-y_{4};

hence scale (h) is the same as scale (m) revolved 180° about the center of the circle.

5. In Fig. 4 the two preceding nomograms are superposed and drawn to a larger scale (making 1.6 cm. instead of 1.4 cm. to show the details of the figure more clearly), and since the problem requires h only, the common scale (m) is left blank. The resulting abacus is a circular one of double alignment and four variables.*

From the preceding we see that if we have given a,) ? and a' to find h, we first join the point a+? of the scale (a+?) to the point ? of the scale (?) by means of a straight line, and thus determine another point where this connecting line cuts the circle [this point would give the value of m, if we had constructed the scale (m)]. Next we join this point to the point a' of the scale (a') by means of a straight line, and the point at which this connecting line cuts scale (h) gives the required value of h.

From the above figure, when a=130°, ?=30°, a'=50°, we find h=1.7

*D'Ocagne, see p. 330.

6. If H falls on the base AB, Fig. 5, our method fails, because the values of in and n, given by (1) and (2), and consequently the values of h, given by (5) and (6), are represented in an indeterminate form, which is shown in the nomogram by the coincidence of the points ? and a+?.

We may then proceed in the following manner: Put

AP/AB=m’ BP/AB=n’

Whence by trigonometry we have

m’=sin ?’/sin(a’+?’)

n’=sin a’/sin(a’+?’)

h=m’ sin a’

h=n’ sin ?’

from which, in both cases,

h= sin a’ sin ?’/sin (a’+?’)

Comparing (16) with (1) we see that to obtain m', given a' and ?', the same nomogram will serve, Fig. 2, that was used to find m from a and ; we need only substitute a' and ?' for a and in reading the scales (a+?) and (?). Next, comparing (18) with (3), it is clear that we pass from the latter to the former by substitutings in a' for tan a'; therefore in the nomogram represented by Fig. 3, instead of the scale (a') given by (15), it is necessary to construct the scale given by

x_{6}=l/1+k_{3} csc a’ y_{6}=o

In the complete nomogram, Fig. 4, this new scale is marked ((a’)) and with it, when a'= 130°, and ?=39°, we find h=1.1.

In Fig. 5 it is assumed that H falls between A and B, but even if this is not so, the method will still be general if by a' and ' the angles BAP and ABP are understood.*

NOTE.—When k_{2}=k_{3}=1, we see that for equal values of at and , we have

X_{3}+x_{6}=l

Thus, with respect to the center of the circle, the scale ((a’)) is symmetrical with scale (?) and therefore need not be computed.

7. Another case, which in reality comes under the preceding special case, is that in which H falls on A or B, Fig. 6. It would be sufficient to proceed as in the above paragraph, assuming a'=90°, or ?=90°.

However, equation (20) reduces to one of the two following forms:

h=tan ‘, or h=tan a’

as do also equations (4) and (3); thus it is sufficient (see Fig. 3) to connect the point 1 of scale (m) with the point ?' or a', of the scale (a'), to obtain h directly.

In Fig. 4 the point ((1)) is also the point 1 of the blank scale (m). From this figure, as also from Fig. 3, when a'=15° we find h=0.26.

8. In the accompanying plate we have assumed greater dimensions for the abacus in order to show it to better advantage. Its arrangement has also been changed to facilitate its use.

The coefficients k_{1}, k_{2}, k_{3}, have been so chosen that scale (h), from 0 to 30, may be as distinct as possible; to this end, after a few trials, it has been found satisfactory to assume k1=0.4 and k2=k3=2.

*The graph of M. Petit is not only restricted to the case discussed in this paragraph, but is also limited by the hypothesis that H falls between A and B. Yet "...son exactitude, tres superieure a celle du barometre, l'a fait adopter a peu pres partout ou? il y a des grandes semaines."

The coefficient l has been taken equal to 20 cm., but with a design four or five times this size we could certainly reach approximations, in portions of the base AB, of 1/100 when h is between 0 and 5, 1/50 when h is between 5 and 10, 1/25. when h is between 10 and 15, and when h is between 15 and 30, and, considering the nature of the problem, especially the instability of the point observed, with a base AB not less than 50 m., this degree of accuracy is more than sufficient.*

From what has been said in Note III to Art. 2, and also in Art. 5, the nomogram can be used in the following manner: Hold a thread at the point a+? with the left hand; then keeping it taut with the right hand, revolve until it passes through the point ?. Then, without releasing it, hold the thread with the right hand at the point where it cut the circle, and keeping it taut with the left hand revolve it again until it passes through the point a'. The final intersection of the thread with the circle will be the point whose scale reading will give the value of h. This would be the process for the general case of Art. 5, and similarly for the cases of Art. 6, and Art. 7.

Thus in the example of Art. 5, viz., given a= 130°, ?=30°, and we find with greater precision (see the dotted lines I) h=1.75.

In the example of Art. 6, viz., given a'=130° and ?'=30°, we find the more exact value (see the dotted lines II) h= 1.12.

And in that of Art. 7, viz., given a'=15°, we find (see the dotted line III) h=0.27.

These examples are sufficient to show how to proceed in every case.

NOTE I.—For the computation of y_{1}, y_{2}, x_{2} and y_{4}, the simplifications indicated in Note I to both Art. 3 and Art. 4 still hold. But those indicated for the computation of scale (a') in Note I to Art. 4, and for ((a')) in the Note to Art. 6, are no longer possible. And if scale (m is not computed, the simplification mentioned in Note II to Art. 4 has no basis.

* A closer approximation can only be deceptive, and in this connection M. Petit (q.v.) wisely remarks, "Appearances of accuracy are attained that are amusing: a recent official record has been announced at 2682 meters... a little more and they would have given centimeters."

NOTE II.—With a large model the empty spaces might be advantageously filled with directions for the use of the abacus Also there would be room enough to write two sets of supplementary numbers for the corresponding scales, and thus do away with the reduction to the first quadrant in those cases where f3 or a+? is an obtuse angle.