Stadium Sightline

7484 /www/nrich/html/content/id/7484/ 1 2011-06-15T14:55:21 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> <ul id="stemLinks"> <li><a href="http://nrich.maths.org/8041">Warm-up</a></li> <li><a href="http://nrich.maths.org/7023">Try this next</a></li> <li><a href="http://nrich.maths.org/5839">Think higher</a></li> <li><a href="http://sport.maths.org/content/how-velodrome-found-its-form">Read: mathematics</a></li> <li><a href="http://en.wikipedia.org/wiki/Stadium">Read: engineering</a></li> <li><a href="http://plus.maths.org/content/teacher-package-mathematical-modelling">Explore further</a></li> </ul> <div> </div> One of the challenges in designing an Olympic stadium is to make sure that spectators can see the event without their views being blocked by the spectators in front. Your task is to design the tiered seating for the stadium. The back of each seat is 80cm behind the back of the seat in front, but raised up so that each row can see over the row in front. Make some sketches showing what the cross-section (side view) of the seats might look like. Will the seats lie on a straight line, or a curve? How steep? What factors do you think you will need to take into account when working out how high each seat needs to be? The Olympic organisers have stipulated that all spectators must be able to see clearly a point 10m in front of the front row of seating: <mdo:image src="sightline1.png"></mdo:image> The spectator in the second row needs to have line of sight to the same point as the spectator in the first row, as seen in the diagram above. Notice that the spectator in the second row needs some extra clearance in order to see comfortably over the first spectator's head. Here is a zoomed-in diagram: <mdo:image src="sightline3.png"></mdo:image> The first spectator's eye-level is 1.2m above the ground. There is an extra 0.2m of clearance from his eye-level to the second spectator's line-of-sight. The second spectator is a further 0.8m away from the point on the pitch. How high above ground level does spectator 2's seat need to be? Now draw a similar diagram with the dimensions and unknowns for spectators 2 and 3. How high above ground level does spectator 3's seat need to be? Finally, imagine there were 40 rows of seating in the stadium. Can you work out the heights above ground level of each of the 40 rows, and hence plot a side view of the seating? It is very helpful to use a spreadsheet to perform the repeated calculations and plot the results. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">This was a tough question! Only one correct answer to this: Joshua, at Chatham Grammar School for Boys, sent us a great solution. It was as follows: Our first step should be to generalise the dimensions shown, for the nth person in the stadium. We shall let the point which must be visible to all be denoted as P. The distance, d, from P to the nth person's seat is therefore given by: $d_n = 10 + 0.8(n - 1) = 10 + 0.8n - 0.8 = 9.2 + 0.8n$. Furthermore, let $\theta_n$ represent the angle between the horizontal and the line of sight of the n-th person. We assume every person has a torso height of 0.8m. Let $x_n$ represent the height above ground level of the n-th person's seat. We can consider the following: $\theta_{n+1} = \arctan\left(\frac{1 + x_n}{d_n}\right)$. This formula is much easier to consider when we break it down step by step. Consider the zoomed in picture from the original problem. Consider the right-angled triangle formed with the perpendicular at person 1's seat extended to the line of sight of person 2. That triangle will enclose the angle between person 2's line of sight and the horizontal, $\theta_2$. Its height is given by adding the overhead clearance (0.2m), the torso height (0.8m) and person 1's chair height ($x_1$). Its base length is given by the distance from person 1's seat to the point P ($d_1$). This is true generally with 1 and 2 replaced by n and n+1, by considering similar diagrams. Therefore by use of the expression $\tan\theta = \mathrm{opp}/\mathrm{adj}$, we arrive at the above general formula. Now we can replace the $d_n$ in the formula with $9.2 + 0.8n$, to give: $\theta_{n+1} = \arctan\left(\frac{1 + x_n}{9.2 + 0.8n}\right)$. The last step now is to generate a recurrence relationship between $x_{n+1}$ and $x_n$, by eliminating $\theta$. We can do this by getting an expression for $\theta_{n+1}$ in terms of $x_{n+1}$. By again studying the zoomed in diagram, we shall get an expression for $x_2$ and then generalise. Consider the triangle formed between the perpendicular at person 2's seat up to their eye-level and the horizontal along to P. This horizontal distance is $d_2$ and the vertical distance is $x_2 + 0.8$. Therefore, by use of the tangent formula: $\theta_{n+1} = \arctan\left(\frac{x_{n+1} + 0.8}{d_{n+1}}\right)$. By eliminating $\theta_{n+1}$ from these two equations, we eventually end up with: $x_{n+1} = \frac{2.64 + 0.16n + (10+0.8n)x_n}{9.2 + 0.8n}$ and we can now set up a table using this formula to calculate all of the values from $x_2$ to $x_{40}$, as we know that $x_1 = 0.4$. The required values are: $x_2 = 0.712$ $x_3 = 1.0388\dots$ $x_{40} = 19.0444\dots$ Fantastic!</mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><h3>Why do this problem?</h3> <a href="http://nrich.maths.org/7484">This problem</a> encourages students to model a real-life situation mathematically, that of tiered seating design in sports stadia. The problem begins with some straightforward geometry but leads into some spreadsheet modelling. Although the problem can be solved using trigonometry, the most elegant solution simply uses similar triangles. <h3>Possible approach</h3> "Your task is to design the seating layout for the stadium. The seats will be tiered so that the back of each seat is 80cm behind the back of the seat in front, but raised up so that each row can see over the row in front. Start by making some sketches of how you think the seats might need to be arranged." Give students time to try out some ideas. Perhaps sketch on the board (or invite students to sketch) a variety of possibilities - straight lines of different gradients, and convex and concave curves. Then allow some time for students to share their intuitions about which will best model a successful tiered seating design, together with any reasons they have to support it. Next, introduce the diagram: <mdo:image src="sightline3.png"></mdo:image> Set the next task, to work out the missing height. Students may wish to draw or plot accurate diagrams before making calculations. Move on to working out the height above ground level of the third seat, and allow some time to discuss whether students are surprised that the extra height is not constant. Finally, invite students to explore what happens for subsequent steps, perhaps using spreadsheets to iterate and calculate the next step height from the previous. <h3>Key questions</h3> What sort of a view would spectators have if each seat was higher than the one before by the same distance? What sort of a view would spectators have if the angle between their line of sight and the line of sight of the spectator in front was constant? <h3>Possible extension</h3> Invite students to consider other factors that might affect stadium seating and perhaps adapt their models accordingly. <h3>Possible support</h3> <a href="/7359">Track Design</a> looks at a simpler design context at the Olympics. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">What lengths do you know? Look for similar triangles.</mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> One of the challenges in designing an Olympic stadium is to make sure that spectators can see the event without their views blocked by others. Suppose the horizontal distance (D) between the spectator and the event is 30m and the vertical height (R) is 10m. What is the viewing angle of the spectator? If the depth of the seating row (T) is 1m, how much should the riser height (N) be such that the view of the spectator does not get completely blocked by the person in front? The spectator's ability to see the event over the head of the person in front is measured by the 'C' value (see the illustration below) <mdo:image width="400" height="260" alt="" src="sightline_eqn.gif"></mdo:image> (Image taken from Paul Shepherd's website: http://people.bath.ac.uk/ps281/) A reasonable number for C is 15cm. To make sure that the spectator can comfortably watch the event, how much should the riser height (N) be? In what other ways can we improve the C value of the spectators? The C value can be calculated using the formula $$C = \frac{D(N+R)}{D+T} - R$$ (Try derive it yourself!) Can you relate everyday experience to the formulae above? Which factor affects the C value most? Suppose the seating area is limited by the overall stadium design to be 20m in seating depth and 10m in vertical height, how can you optimize the viewing comfort while accommodating as many spectators as possible? </mdoxml> 2 3 0 0 0 1 1 Stadium Sightline How would you design the tiering of seats in a stadium so that all spectators have a good view? Applications sport Trigonometry Trigonometry generally 3D Geometry, Shape and Space Plans & elevations 2D Geometry, Shape and Space Similar triangles Admin Individual Applications DT Applications Technology Applications Maths Supporting SET Admin Featured STEM