Blake Cicenas
Physics 710
Capacity
To start with the lawful capacity problem, I created a zeroth order model based on a few assumptions. I assumed that the room would be rectangular of dimensions l x w and that the flow of people would be smooth as they would all flow together with the same velocity; essentially a fluid-like flow but made up of people. Next, I began to look for the time it took for the last person to exit the room. I figured that the capacity problems are based on evacuations and I should really only worry about the time it takes for the last person to leave the classroom. The classroom set up is very similar to the room Physics 710 takes place and I assume that there were 2 square desks in the room of dimensions D x D horizontally spaced an average distance i across. The desks were placed a distance H from the front of the room to allow the lecturer to walk about the room without bumping into desks. Finally, it is assumed that the last person at the left hand corner will stand up and walk to the front of the room, then to the left to clear all of the desks before turning self again and heading down to the exit. A picture of the simple situation is contained below.
With these assumptions a time t can easily be found for the time it takes to get the bottom right person out of the room at exit e. The equation becomes which can then be rewritten as where nyis the number of horizontal rows of desks and nxis the number of vertical columns of desks. This is done to account for extra desks and it will be assumed that there will always be an even number of desks in the room. Also, a 2 was included at the front to account for walking up to the front of the room and walking back. Now, if I factor in the average vertical distance iybetween the desks the equation will then become where I was rewritten as iy and ix to describe the desk spacing both horizontally and vertically. The equation can then be rewritten as . However, I have neglected both the length and the width of the room. I cannot yet work them into the equation although I can set up the parameters as and . Note that I also included the distance H because more than likely a lecturer would like to have room between the desks to teach and lecture. Now I can also note that to maximize space I could set x equal to zero since the width of the door should provide enough room for people to walk. However, I will keep that in mind and come back to that at a later time.
Now I will include the fact that the desks are rarely square and are more than likely rectangular. So I will redraw the picture to include rectangular desks as well as extra rows and columns. This will look like…
This picture also alludes to the fact that in general, H must be greater than or equal to D so that the instructor has enough room to move around. Also to note is the bottlenecking due to the fact that successive rows must wait for the previous row to exit, quite similar to Catholics exiting their respective pews for communion. With this in mind, I must determine how long it takes a person to cover a set distance. So I will assume that the distance between the last person of the leading column to the last person of the lagging column will be (Dx+ix). I also know that the wait will be based on the ny, or the number of rows that occur. However, I have to keep in mind that I am only concerned about how long it takes to get the last human at the far right corner out of the room. And the wait for the final human will be based on both ny and nx, or the number of people and desks that are in the room. So I will define the waiting time to be twand will find an equation to represent this time. I can assume that the first column will have no waiting time while the following columns will each have longer and longer lag times. Another way of putting it is that the last person in the second column will have to wait 1 ny to travel a distance of . And this process will occur (nx - 1) times where the minus one indicates that the first row is independent of waiting. Also each successive row will have to travel a distance an extra distance that can be described by (Dx+ix)(nx - 1). The equation can then be solved for twto get
I can then describe the total time as ttot = tL+twwhere ttot is the total time and tL is the time for the last person to exit. Substituting the values in yields
where the parameters for the equation are and
Now that I have a very basic equation, I should try to factor in the room dimensions. Since for all purposes they are equal, I can sayand Thenand multiplied out is
Interesting to note is that (DxDy) is the area of the desk and (nxny) is the number of the desks. This set-up is handy and could aid me later in finding the optimum number of desks for a given room area.
Now I will attempt to see if the model works. I start with defining some of my variables. I will assume a desk is 2.5 ft wide (in the x –direction) and 3.5 ft long(in the y – direction) so Dx=2.5ft and Dy=3.5ft. I will leave 6 inches between the desks in their respective columns and an isle space of 3 ft so iy = 0.5ft and ix = 3 ft. I will also assume the minimum velocity would be 2 mph which corresponds to 2.9 ft/s. I will also let H = 12 ft which should allow enough room for a desk and to move around and teach. I will also assume the door is 4 ft across with an initial walkway of 3 ft so e = 4 ft and x = 3 ft. Finally I’ll say I have 24 desk in 4 columns of 6 desks so nx = 4 and ny = 6. So the equation then becomes which yields 77 seconds. This time is under our very simple “2 minutes” that we decided as an arbitrary time to escape the room.
My next step is to star to research what the good times actually are and to start to incorporate more complicated situations into the model such as an odd number of desks coupled with the fact that a “Catholic Communion” exit strategy will not necessarily be the way people will evacuate a room. My research so far has turned up a few websites where you punch in numbers and they give you the maximum number of people that can be in the room based on a few parameters. However, I did find a paper that details how long it takes to evacuate a particular room based on some particular parameters. But most importantly this paper has codes at the end and can be found at . I will spend my next week looking at that as well as still examining the thermodynamic paper since it has applications to this problem.