On Growth and Form

by D'Arcy Wentworth Thompson

Fig. 145.

we may deal with the case as though we had now to do with four equal cells, each one of them a perfect quadrant. And so our problem is, to find the shortest line which shall divide the quadrant of a circle into two halves of equal area. A radial partition (Fig. [145], A), starting from the apex of the quadrant, is at once excluded, for a reason similar to that just referred to; our choice must lie therefore between two modes of division such as are illustrated in Fig. [145], where the partition is either (as in B) {360} concentric with the outer border of the cell, or else (as in C) cuts that outer border; in other words, our partition may (B) cut both radial walls, or (C) may cut one radial wall and the periphery. These are the two methods of division which Sachs called, respectively, (B) periclinal, and (C) anticlinal[391]. We may either treat the walls of the dividing quadrant as already solidified, or at least as having a tension compared with which that of the incipient partition film is inconsiderable. In either case the partition must meet the cell-wall, on either side, at right angles, and (its own tension and curvature being everywhere uniform) it must take the form of a circular arc.

Now we find that a flattened cell which is ap­prox­i­mate­ly a quadrant of a circle invariably divides after the manner of Fig. [145], C, that is to say, by an ap­prox­i­mate­ly circular, anticlinal wall, such as we now recognise in the eight-celled stage of Erythrotrichia (Fig. [144]); let us then consider that Nature has solved our problem for us, and let us work out the actual geometric conditions.

Let the quadrant OAB (in Fig. [146]) be divided into two parts of equal area, by the circular arc MP. It is required to determine (1) the position of P upon the arc of the quadrant, that is to say the angle BOP; (2) the position of the point M on the side OA; and (3) the length of the arc MP in terms of a radius of the quadrant.

• (1) Draw OP; also PC a tangent, meeting OA in C; and PN, perpendicular to OA. Let us call a a radius; and θ the angle at C, which is obviously equal to OPN, or POB. Then
• CP = a cot θ; PN = a cos θ;
  NC = CP cos θ = a · (cos² θ) ⁄ (sin θ).
• The area of the portion PMN
• = ½ C P² θ − ½ PN · NC
  = ½ a² cot² θ − ½ a cos θ · a cos² θ ⁄ sin θ
  = ½ a²(cot² θ − cos³ θ  ⁄ sin θ).
• {361}
• And the area of the portion PNA
• = ½ a²(π ⁄ 2 − θ) − ½ ON · NP
  = ½ a²(π ⁄ 2 − θ) − ½ a sin θ · a cos θ
  = ½ a²(π ⁄ 2 − θ − sin θ · cos θ).
• Therefore the area of the whole portion PMA
• = a² ⁄ 2 (π ⁄ 2 − θ + θ cot² θ − cos³ θ ⁄ sin θ − sin θ · cos θ)
  = a² ⁄ 2 (π ⁄ 2 − θ + θ cot² θ − cot θ),
• and also, by hypothesis, = ½ · area of the quadrant, = π a² ⁄ 8.
• Fig. 146.
• Hence θ is defined by the equation
• a² ⁄ 2 (π ⁄ 2 − θ + θ cot² θ − cot θ) = π a² ⁄ 8,
  
• or
• π ⁄ 4 − θ + θ cot² θ − cot θ = 0.
  
• We may solve this equation by constructing a table (of which the following is a small portion) for various values of θ.
• θ
• π ⁄ 4
• − θ
• − cot θ
• + θ cot² θ
• = x
• 34° 34′
• ·7854
• − ·6033
• − 1·4514
• + 1·2709
• =  ·0016
• 35′
• ·7854
• ·6036
• 1·4505
• 1·2700
• ·0013
• 36′
• ·7854
• ·6039
• 1·4496
• 1·2690
• ·0009
• 37′
• ·7854
• ·6042
• 1·4487
• 1·2680
• ·0005
• 38′
• ·7854
• ·6045
• 1·4478
• 1·2671
• ·0002
• 39′
• ·7854
• ·6048
• 1·4469
• 1·2661
• − ·0002
• 40′
• ·7854
• ·6051
• 1·4460
• 1·2652
• − ·0005
• {362}
• We see accordingly that the equation is solved (as accurately as need be) when θ is an angle somewhat over 34° 38′, or say 34° 38½′. That is to say, a quadrant of a circle is bisected by a circular arc cutting the side and the periphery of the quadrant at right angles, when the arc is such as to include (90° − 34° 38′), i.e. 55° 22′ of the quadrantal arc.
• This determination of ours is practically identical with that which Berthold arrived at by a rough and ready method, without the use of mathematics. He simply tried various ways of dividing a quadrant of paper by means of a circular arc, and went on doing so till he got the weights of his two pieces of paper ap­prox­i­mate­ly equal. The angle, as he thus determined it, was 34·6°, or say 34° 36′.
• (2) The position of M on the side of the quadrant OA is given by the equation OM = a cosec θ − a cot θ; the value of which expression, for the angle which we have just discovered, is ·3028. That is to say, the radius (or side) of the quadrant will be divided by the new partition into two parts, in the proportions of nearly three to seven.
• (3) The length of the arc MP is equal to a θ cot θ; and the value of this for the given angle is ·8751. This is as much as to say that the curved partition-wall which we are considering is shorter than a radial partition in the proportion of 8¾ to 10, or seven-eights almost exactly.

θ	π ⁄ 4	− θ	− cot θ	+ θ cot2 θ	= x
34° 34′	·7854	− ·6033	− 1·4514	+ 1·2709	=  ·0016
35′	·7854	·6036	1·4505	1·2700	·0013
36′	·7854	·6039	1·4496	1·2690	·0009
37′	·7854	·6042	1·4487	1·2680	·0005
38′	·7854	·6045	1·4478	1·2671	·0002
39′	·7854	·6048	1·4469	1·2661	− ·0002
40′	·7854	·6051	1·4460	1·2652	− ·0005

Fig. 147.

But we must also compare the length of this curved “anticlinal” partition-wall (MP) with that of the concentric, or periclinal, one (RS, Fig. [147]) by which the quadrant might also be bisected. The length of this partition is obviously equal to the arc of the quadrant (i.e. the peripheral wall of the cell) divided by √2; or, in terms of the radius, = π ⁄ 2 √2 = 1·111. So that, not only is the anticlinal partition (such as we actually find in nature) notably the best, but the periclinal one, when it comes to dividing an entire quadrant, is very considerably larger even than a radial partition.