   # Nomography - Encyclopedia

Putting v=o, u = u =0, v = xfl+ygi +hl =o xf2+yg2 +h2 o xf 3 +yg3 +h3 =o Fig: 10.

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-40 -30 _20 x - X Ai - A2 The expression for x being constant we see that the points are arranged along a straight line Cw parallel to Au, B y, and this straight line cuts AB at a point C such that CA___ µ1 CB µ2 Then writing for the scale (z 3) on Cw w = as w is the same as y we have f:43 or = + - /23 /21 In the particular case where /11 = /22 we will have /2l 2 2 To recapitulate, the practical procedure is shortly as follows: - Select suitable axes Au, B y and suitable units /21, . Draw Cw dividing AB in the ratio, and determine the unit /13 by the relation I I I ' ' - Construct along Au, B y, Cw, respectively the scales =µ1f1 Any straight line drawn across these three will then cut them at cor responding values of z 1, z 2, z 3 as de fined by (2).

As a rule we arrange the diagrams so that the scale of the variable, which (Z,) (_3) (,2) has generally to be determined in terms of the other two, lies between their scales, as this conduces to greater accuracy in reading.

It is not necessary for the origins A, B, C, to appear on the diagram unless they are required for the range of the variables for which the formula is to be employed. The scales can be quite easily constructed without them, starting from the lowest value required. It will be seen that the freedom of choice of axes and units renders this method an exceedingly flexible one. Examining the range of the variables required for the practical use of the particular formula, we can arrange the scales and the size of their graduations to the best advantage.

Among other things, we wish to avoid the reading straight line making too acute an angle with the scales, as this leads to inaccuracy. Practice will soon enable the best disposition to be seen, but it will most frequently be found convenient to make the useful parts of the scales (z 1), (z 2) about the same length and about the same distance apart, so that the complete diagram is roughly contained in a square.

As an example of Type A, the formula d= ,,1/K2 already referred to in §2 can be taken, supposing that it is now desired to construct a nomogram to show different values of K, instead of a single curve for a constant value of K.

The formula can be reduced to Type A by writing it log d+3log K - log C=o and taking --d, f 1 = log d = K, f2=; log K = log C the scales are all logarithmic scales, differing only as regards their unit.

Take any convenient logarithmic scale that may be available (say that of a slide rule) and by means of it graduate the scales (d) and (K) on two convenient parallel axes (fig. 14).

We can then determine the point C = to on the scale (C) by the cross alignments d= l, K=to d=2.5, K=80 for both of which C = ro.

The support of (C) is then a straight line parallel to the axes through this point, and we can graduate it by noticing that for C = K, we always have d= I .

The alignment of d= 1 with K =20, 30, 40, 50, in turn, then gives the points C = 30, 40, 50,....

Suppose now that we want to know the current which will fuse an aluminium (K =59) wire o3 mm. diameter. The straight line joining o3 on the (d) scale to 59 on the (K) scale (see dotted line, fig. 14) cuts the (C) scale at about 9.5 amp., the required current.

 i

## Type B

Nomograms with three rectilinear scales, two of which are parallel. If the formula to be represented can be put in the form f1.-f-f2 h 3 - o (6) it can be represented by two systems of points (z i), (z 2) arranged along two parallel straight lines, and a third system arranged along a straight line making an angle with the other two.

As before, we take the functional scales u =/2l =µ2f2 along the parallel axes Au, B y (fig. 25).

We then have for the system (23) I 2.2 u + / 21 h 3 y =o which with our usual axes defines the system of points x /21h3 = - /22 O i, y= h so that the points of the system are arranged along AB.

The scale (z 3) can be graduated by the use of the above expression for x, or from a double-entry table of corresponding values of z l, z2, z 3, by suc cessive alignments of pairs of values of z l, 2 2 corresponding to any graduation z 3. Thus if a and c are a pair of values of 2 1 and z 2 which correspond to the value b of z 3, we join ac to cut AB at b, which gives the graduation of the scale for the value b. If A and B do not appear on the diagram the support of the scale (z3) can be drawn by making use of the relation, that if 6 1, 6 2 are the distances of a point on (z 3) from Au, By, we have 62 62 On the completed diagram any straight line drawn across the three scales will cut them at corresponding values of z 1, 22, z 3 as defined by (6).

The scale (z 3) will lie between or outside the scales (z 1), (z2) according as to whether h 3 is positive or negative, and, as in the previous type, it is as a rule best to arrange that the scale of the variable, which generally has to be determined in terms of the other two, lies between their scales; h 3 can always be made positive or negative as desired, altering if necessary the signs of both f 2 and h3. As an example of Type B take Sir Benjamin Baker's Rule for the weight of rails W= 17:,/ (L + Ly2)2 where L=Greatest load on one driving wheel in tons.

y =Maximum velocity in miles per hour.

W.-Weight of rails in lb. per yard.

 8
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Writing the formula I+0 00011,) 7)3-0 z1 =L, f1 = L =V, f2= 1+000010 - W)3 l construct the scales u= /2lf =/21 L = and taking u The support of the scale (W) is the straight line joining the zero of the (L) scale to the zero of the f2 scale. This latter zero is at an inconveniently great distance from the top graduation on the (v) scale, but the support can readily be obtained without the actual use of the zeros of the f i and f 2 scales by the use of the formula al referred to above, or by a cross alignment in the following way: - Take W = loo and work out L for v =60 and 80. The straight lines joining these two values of L and v will intersect at the point 200 on the W scale. Joining this point to the zero of the (L) scale gives the support, and the remainder of the scale can be graduated by taking v = 50 say, and working out L for W = 20, 30, 40, 50.... Joining 50 on the (v) scale to fLI these values of the (L) scale in 10 tons turn, will give an intersection on pi the support for the corresponding graduations of the (W) scale.

The completed diagram is shown in fig. 16. To use it sup pose, for instance, we require the / sovalue of W for L =7 tons, v = 70 m./hour. The straight line join ing 7 on the (L) scale to 70 on s / o sothe (v) scale (see dotted line, fig.

16) cuts the (W) scale at about 82 lb./yd., the required value.r 4 w-rr, 7 (L Type C. - Nomograms with two parallel rectilinear scales and one 130 '30- ' curvilinear scale. If the formula to be represented can be put Ito loin the form (7) it can be represented by two systems of points (z1), (22), arranged along two parallel straight lines, and the third system (z 3) arranged along a curve.

As in the preceding types we take functional scales =µi fs u =µ2f2 along the parallel axes Au, B (fig. 17). We then have for the system (23) g which with our usual axes defines the system of points x - 1 _ h + y _ /Llh3 8 and we ca determine any number of points on the system (23) by means of these equations, or by a series of cross alignments. This latter method is especially indicated in cases in which a double-entry table of corresponding values of z l, 22, 23, is already available.

Proceeding by whichever of these ways is most convenient, we can obtain the complete scale (z3), tracing the curvilinear support through the points determined.

As before it is advantageous, where the variable z 3 is generally the unknown, for the scale (z 3) to lie between the scales (z1), (22). This will be the case if g3 is positive, and this can always be arranged, if necessary, changing the signs of both f 2 and h3.

Having constructed the scales (z l), (22), (z 3) as described above, any straight line drawn across them will cut the three scales at corresponding values of zl, z2, 2 3 as defined by (7).

As an example of Type C take the formula used for the thickness of cast-iron pipes in waterworks, t =o000 125 P d -F- 0 15 'VT where t = Thickness of metal in inches.

P = Pressure of water in pounds/inch.

d =Internal diameter of pipe in inches.

Writing this, 0 000125 P d - t -}- 0.15 o and putting z1 f1=P z3 =d, f 3 =0 15 1/71, g3=d, h8= we see that it is of type C.

We construct the scales u =µ1P y=/22t along two convenient parallel axes (fig. 18).

We then determine sufficient points on (d), by cross alignments, to draw the curve and graduate the scale.

When P is zero, t =0.15 ¦d giving us an easily calculated series of alignments for d = 5, to, 15... For the cross alignments it will be convenient to take P = loo, and calculate t for d = 5, 10, 15, ... as before.

producing the straight line to cut the (t) scale (see dotted line, fig. 18), we get, at the point of intersection, the required value t = P3 in.

7 Graphic Representation of Formulae with more than Three Variables. (i.) Double Alignment Nomograms. - Certain types of formulae containing four variables can be dealt with by breaking them up into two or three variable formulae with a common auxiliary variable.

Consider for instance a formula which can be written in the form fl (8).

Introducing an auxiliary variable z 5 we can construct two partial nomograms f1-1112 (9) f3 +f (10) 7t':: of Type A, having the scale (25) in common.

Such an arrangement is shown schematically in fig. 19, the central line representing the auxiliary scale.

If we take values of 21, 2 2, 2 3, 2 4 satisfy-, ing equation (6), the alignment of z 1 with 2 2, and of z 3 with 2 4 will intersect on the scale (25). The central line need not be c--" d graduated as it is only required as a refer- ence line, the nomogram being read in the following way: Suppose we require the value of z 4 for z1= a, 2 2 =b, z 3 = c. Join a on (z l) to b on (22) cutting the reference line at e. Join c on (23) to e and produce to cut (24) in d, which will give the corresponding value of 24.

Such a nomogram from the way in which it is read is termed a Double Alignment Nomogram.

It will be noticed that as the unit of (25) is the same in both (9) and (to) we must have the relationship I I I I /14 while the distances of the scales from the reference line (fig. 19) will be given by 63 62 112' 64 Hence for the practical construction we graduate any of the three scales, (21), (z 2), (23), say, from three conveniently chosen origins on the supports of their scales. We then determine a point on the scale (24) by means of four values of z i, 2 2, 23, 2 4, (a, b, c, d, say) which satisfy equation (8).

The alignment of c and the intersection of the alignment ab with the reference line then determine the point d on the scale (z4)

and as we know /14 we can construct the scale (24) completely. As an example take the formula for the discharge of gas in pipes, Q - 1050 D"- N 1 H 0.45L where L =Length of pipe in yards.

D = Diameter of pipe in inches.

H =Head of water in inches equivalent to the pressure.

Q =Quantity of gas discharged in cub. ft. per hour. Writing it log log L = 2 5 log D +1 log H -{- const. We put z1=Q, z2 L, 23=H, z4=D.

 0-
 -` _ look 0 c?J ?,: - so 80 1.6- d

along two convenient parallel axes. The scale (L) is an evenly divided scale, and to graduate the scale (v) a series of values of v and f2 are calculated v = 20 30 40 50 60 708012-0961 - 09171-0862 - 0 800 - 0.735 - 0 671 - 0.610 f Ig34-f2h3+f =0 Suppose now we wish to know the thickness of a pipe of so-in. bore to stand a pressure of 130 lb./in. Joining 130 on the (P) scale to 30 on the (d) scale, and 40- u _0 Z Z (Z4) 0. 62 6, 63 19.

Fig. shows the resulting nomogram constructed with µl=2/12=2µ3=µ4 ioo-yd. pipe of 1-in. bore, with a head of water of Suppose now that we want to know the rate of discharge from a I in.

 2610.5 (Q) _500o - _100 (0)

Fig: 20.0' 6 Join 1 on the (D) scale to 1 on the (H) scale. Join the point where this straight line cuts the reference line to ioo on the (L) scale, and produce the straight line backwards to cut the (Q) scale at 201 ft./hour (see dotted lines, fig. 20), the required rate of discharge.

These Double Alignment Nomograms can be constructed by combining any two of the types, A, B, or C where the four-variable formula can be written in the appropriate form.

Take for instance the formula H = 18,400 (log B 1 - log B 2) (I + 0.003670) giving the difference in level (H metres) between two stations at which the barometric readings are B 1 and B2 mm. respectively, the mean temperature being 0°C. = t 1 2 t2) Writing it B and A 600 Fig. 21 shows the resulting nomogram, and, to illustrate its use, suppose B 1 = 750 mm., B2 = 670 mm. and it is required to find H.

Join 670 on (B 2) to 750 on (B 1), and produce to cut the reference line. Join the point thus obtained on the reference line to on (0). This straight line produced will cut (H) at 935 mm. (see dotted line, fig. 21), the required difference in height.

(ii). Combination of an Alignment Nomogram with a Network. Suppose we have a network (z 1, z 2), composed of two systems of figured curves (z5), (72) crossing each other (fig. 22).

If we take any point on this network, a curve of both systems will pass through this point, and we may assign to the point a value of both z i and of z 2, taking the values from the curves of the systems (z 1), (z 2) which intersect in the point. The point has thus in a sense two values and is termed a binary point. The general equation in parallel coordinates of such a binary point will be of the form 112+g12u±h12v =o and its coordinates in cartesians with the usual axes will be = h12 - g12 = µf12 x - Xh g12 y h12 + We can obtain the equations of the systems (z 1), (72) forming the network (z 1, 7 2) by eliminating in turn z 2 and z 1 between the above expressions for x and y. Consider now a formula that can be put in the form fig34+f2h34+f 34 =0 This can be represented by the rectilinear parallel scales u=Alf' v=µ2f2 and the network µ2g34u?µ1 +µ1µ2 34 0 or µ1h34 µ2g34' y µ1h34+ µ2g34 As an example take the Compound Interest Formula M =PR" where P is the principal, M the amount, R the amount of I for one year at r % per annum (i.e. R = 1 -} n the number of years. Writing it and taking z l =P, f 1 =log P z 2 =n, 2=n z3=r, f = - log M z 4 = 111, g34 = I, h34 = log R the nomogram will consist of the parallel scales u = / 1 1 log P v =µ2n and the network (r, M) defined by µ2u +µ1 log Rv - µ1µ2 log M =o or in cartesians µl log R-112, _ µ1 µ2 log M Al log R+ /12 y /11 log The expression for x is independent of M, so that we have for (r) a system of straight lines parallel to (P) and (n).

For the system (M) we have, eliminating R between the above expressions for x and y, 2Xy = / L i log M (5 - x) hence (M) consists of straight lines radiating from the point x=X, y =o (i.e. the zero of the n scale), and cutting the straight line x= - X (i.e. the P scale) at the points y= - A i log M so that the lines of the system (M) are easily drawn from the graduations of (P).

Fig. 23 shows the completed diagram. Suppose, for instance, we want to know the amount of £300 in 10 years at 5% compound interest. Joining 3 on the (P) scale to 10 on the (n) scale, this straight line cuts the 5% line (see dotted line, fig. 23) at a point corresponding to the line £490 of the system (M).

Bibliography. - M. D'Ocagne, Traite de Nomographie (1899); Calcul Graphique et Nomographie (1908); Principes usuels de Nomographie avec Application a divers Problbmes concernant L'Artillerie et L'Aviation (1920); Lt.-Col. R. K. Hezlet, Nomography (1913); J. Lipka, Graphical and Mechanical Computation (1918); C. Runge, Graphical Methods (1912). (R. K. H.) -8 6, P 2 5 ?

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