Howzat,... for a New Take on Run Outs in Cricket? Author(S): Elizabeth M
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Howzat,... for a new take on run outs in cricket? Author(s): Elizabeth M. Glaister and Paul Glaister Source: Mathematics in School, Vol. 44, No. 2 (MARCH 2015), pp. 37-41 Published by: The Mathematical Association Stable URL: https://www.jstor.org/stable/24767726 Accessed: 06-07-2021 16:00 UTC JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range of content in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new forms of scholarship. For more information about JSTOR, please contact [email protected]. Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at https://about.jstor.org/terms The Mathematical Association is collaborating with JSTOR to digitize, preserve and extend access to Mathematics in School This content downloaded from 86.59.13.237 on Tue, 06 Jul 2021 16:00:36 UTC All use subject to https://about.jstor.org/terms How2a*,How*at, ...... for for a new a takenew on runtake outs on in cricket?run outs in cricket? byby Elizabeth Elizabeth M. Glaister M. and Glaister Paul Glaister and Paul Glaister DuringDuring four fourdecades ofdecades listening toof Test listening Match Special to Test Match Special cricketcricket commentaries commentaries we have often we heard have the phrases often heard the phrases "he"he has has only onlytwo stumps two to aimstumps at" or "he to could aim only at" see or "he could only see oneone and and a half a stumps half when stumps he let go when of the ball".he Theselet go of the ball". These refer to the scenario when a fielder aims to hit the wicket to run out a batsman while he is attempting to make his ground in the process of taking a run. The most satisfying run out we have witnessed was when Ricky Ponting, the Australia captain at the time, was run out by substitute fielder Gary Pratt in the infamous 2005 Ashes series (Note 1). Four years later Andrew Flintoff ran out Ponting (still Australia captain), in an equally spectacular fashion (Note 2). In this article we examine how we might interpret these phrases from a o o o mathematical point of view using some simple geometry Fig. 1 and trigonometry. Clearly the more stumps that are visible the easier it is to hit them. Figure 1 shows a typical scenario where the fielder upper tangent to the left-hand or leg stump and the lower making the throw is located at P and the three stumps, tangent to the right-hand or off stump. If the distance at least one of which has to be hit, are also shown. between these two tangent lines is nd then we say that n The stumps must have a diameter, d, in the range stumps are visible and shown as the distance AB in 3.49 cm < d < 3.81 cm, and the total width of the wicket, Figure 3. This is the width of the shadow of the wicket w, which is the distance between the edge of the two when the centre of the middle stump is directly between outer stumps along the line through their centres, has the sun and the fielder. In all cases we have located the value w = 22.86 cm. This means that the distance fielder at P close to the stumps so that the figures are between the centres of the stumps, denoted by a, clearer.is given by a = j(w-d), for values between 9.525 cm and 9.685 cm, Our aim is to determine the relationship between the since, as shown in Figure 2, w = 2a + \d + \d = 2a + d. number of stumps, n, visible, and the position of the We define the number of stumps, n, visible to the fielder fielder relative to the wicket, by which we mean the as follows. In Figure 3 we have drawn the line angle,through Qn, that the line from the fielder to the centre of the centre of the middle stump to the fielder atthe P, middle and stump makes with the line through the the two lines that are parallel to this line that centresare the of the stumps, as shown in Figure 3. w © Fig.Fig. 2 2 Mathematics in School, March 2015 The MA website www.m-a.org.uk 37 This content downloaded from 86.59.13.237 on Tue, 06 Jul 2021 16:00:36 UTC All use subject to https://about.jstor.org/terms a *d Fig. 3 We first look at the special cases of three, and then two In the case of two stumps being visible, i.e. n = 2, then stumps being visible, before considering how to the distance between the tangents shown in Figure 5 is approach the general case. In the case of three stumps 2d. In this case the fielder will be located on the line being visible, i.e. n — 3, then the distance between the through the centre of the middle stump which makes an tangents shown in Figure 4 is 3d. In this case the fielder angle 02, with the line through the centres of the stumps. will be located on the line through the centre of the This time the lower tangent to the leg stump passes middle stump which makes an angle 03 with the line through the centre of the middle stump and is the upper through the centres of the stumps. Here we see that the tangent to the off stump. From triangle OTM we see that lower tangent to the leg stump is the upper tangent to . OT \d d d 1 sin 8, = = —= — = = . (3) the middle stump. Similarly the lower tangent to the OM a 2 a w—d w/d-l middle stump is the upper tangent to the off stump. w Note from (1) and (3) that sin03 = 2sin0,. Taking — = 6 From triangle OSM we see that as above we have . OS \d+\d d d 2 ... sin0, = = - — = - = = • (1) sin02 = ^ and 02 = 11.5° (4) OM a a \{w-d) w/d — l as the angle the fielder makes with the line through the Taking the value for the diameter of the stumps centres as of the stumps when two stumps are visible. d = 3.81 cm (at the upper end of the allowable range), Note that a further special case is when only one stump is and the total width of the wicket w = 22.86 cm, then visible, i.e. n — 1, and the distance between the tangents shown in Figure 6 is d. In this case the fielder will be = 6, so that: sin03 = ^ and 03 = 23.6° (2) located on the line through the centres of the stumps, as the angle the fielder makes with the line through the which clearly makes an angle 0, = 0° with the line centres of the stumps when three stumps are visible. through the centres of the stumps, as shown in Figure 6. "p\ \Sd ■h,-' :;k$G Fig. 4 38 Mathematics in School, March 2015 The MA website www.m-a.org.uk This content downloaded from 86.59.13.237 on Tue, 06 Jul 2021 16:00:36 UTC All use subject to https://about.jstor.org/terms B \*t Fig. 5 (•) (•)*(•)"'(•) (•) Fig. 6 We now consider two more special cases case inthe preparation fielder will be located on the line through the for handling the general case. Suppose that centre there of theare middleone stump which makes an angle 02 5 and a half stumps visible, i.e. n = 1.5, with then the the line distance through the centres of the stumps. This between the tangents shown in Figure time 7 is the 1.5d. lower In tangentthis to the leg stump passes through case the fielder will be located on the the line midpoint through between the the centre and the upper part of centre of the middle stump which makes the circumference an angle 0j 5 of the middle stump. The upper with the line through the centres of tangent the stumps. to the Thisoff stump passes through the midpoint time the lower tangent to the leg stump between passes the centre through and the lower part of the the midpoint between the centre and circumference the lower of part the middle of stump. From triangle OVM the circumference of the middle stump, we seeand that also through the centre of the off stump. The upper tangent to the off sin 0 =OV=!£±I^M stump passes through the midpoint between 25 OM the acentre and the upper part of the circumference of the middle stump, and also through the centre of the_ 3 dleg _ Mstump. _ 1 From triangle OUM we see that 4a 2 (w-d) w/d — 1 sinei5 = 2£ = i^M = A = Note— from= ^L_.(1), (3), (5) and(5) (7) that OM a 4 a 2 (w-d) w/d-l sin03 = |sin025 = 2sin02 = 4sin0j 5. Now note that from (1), (3) and (5) we have w w . 3 sin03 = 2sin02 = 4sin015. Taking — = 6Taking again this gives — = 6 again we have sin 02 5 = — and 02 5 = 17.5° (8) ~d as the angle the fielder makes with the line through the sin0.5 10 =L5 — and 0,5 » 5.7° (6) centres of the stumps when 2.5 stumps are visible. as the angle the fielder makes with the line through the We now turn to the general case.