mcm优秀论文.doc

Team # 281 Page 16 of 16 An Energy Model for the Stunt PersonAssumptions:. The total mass (combined weight) of the stunt person (65kg) and the motorcycle (135kg) is 200kg . The motorcycle and the stunt person (MASP), are viewed ideally as a rectangular solid, 1.8m long, 0.5m wide and 1.1m high. To guarantee safety, the effect of 4g is the maximum limit that is allowed. The elephant is viewed ideally as a rectangular solid , 6m long,1m wide and 4m high. It is reported that , to avoid injury, the vertical velocity of MASP cant be larger than 6 m/s while reaching the ground4.Notations(the letters in parentheses are the units of variables):d the width of a box h the height of a box.hmax the altitude of the top point of MASP jumping over the elephant. l the length of a box.L the perimeter of a box (m).P the maximum compression strength of a box (N).Pm the surfacing strength of paper (N/m).Vx the horizontal velocity of MASP.Vy the vertical velocity of MASP.Vs the maximum vertical velocity of MASP reaching the ground safely( 6m/s in our model).W work.The Track of MASPThe Ideal Model -Ignoring Air ResistanceThe Track is shown as Figure 1: Figure 1. MN is a slope to accelerate the velocity To jump over the elephant successfully, the altitude of point A and B of the track must be no smaller than h, that is:HA = HB = where is the angle between the velocity and ground. Given an ,we can get the minimum velocity of v (vmin) and the distance between point D and E ,which is LDE = .The Air Resistance Model-Taking air resistance into accountThe air resistance has significant effect on the motion of MASP and can not be ignored in this model since the MASPs velocity is large. Based on the above analysis and conclusions, we can now take the air resistance into consideration and do the similar analysis conveniently.The track is shown as Figure 2:Figure 2. MN is a slope to accelerate the velocitySince the gravity is much larger than the vertical air resistance, we only focus on the horizontal one. The air resistance is the quadratic function of velocity, thats F = k v2, where coefficient k is a constant5. Based on this conclusion, we get the following equations: For the same principle related before, the altitude of point A and point B must satisfy:HA=HB=We can also get: LDE= For further analysis, please see Appendix A.The Research of BoxThe Maximum Compression Strength of a BoxAccording to reference 1, the maximum vertical strength a box can sustain is P=10Pm L C, where C is a constant and determined by the altitude h of a box.Unfortunately, we get little information about the relationship of C and h or an empirical formula, which we have to do by ourselves. From the data given in reference 1, we can determine the value of C corresponding to specific h (Figure 3), which is represented as vector ( h , c). The data are as follows:(0.165, 5.17583) (0.190, 5.00570) (0.230, 4.65409)(0.245, 4.60045) (0.304, 4.04274) (0.310, 4.00022)(0.335, 3.86207) (0.350, 3.69107) (0.365, 3.62987)(0.380, 3.50039) (0.402, 3.30092) (0.450, 2.83309) (0.510, 2.34030) (0.545, 2.05601) (0.605, 1.56072) From observation, we can approximate c with a linear function. Using the Least-Square Method, the equation is fitted asC = C(h) = C1+C2 h = (6.459-8.0798h)10-2, The boxes in our model are under vertical and horizontal pressure simultaneously, so using the vector addition law, the compression strength is modified to:P=20Pm.Figure 3 Compression Strength of a Folding BoxThere are three stages in the course of compression of a box, i.e. ,elasticity phase ,plasticity phase and collapse phase. If the deformity of a box is x, then the elasticity force F is 2:F(x) = a x e-bx ,where a and b are constant and to be determined.When the deformity of the box is x0=,F attains its maximum value, which equals to the maximum compression strength P. By F(x0) =0 and Max F(x) = F(x0) =P,we get The Energy that A Box Can AbsorbThe total energy that a box can absorb is:W =W can be divided into Wv and Wh , representing the vertical and horizontal absorbed energy respectively:Wv = WWh = WConstructing the Model The Energy ModelIn essence, the effect of cardboard boxes to cushion the fall is to eliminate the energy of MASP. To avoid injury, the vertical velocity of MASP while reaching the ground cant be larger than VS =6 m/s. According to the energy conservation law, the decrement of the energy of MASP is the increment of the energy of boxes, ignoring the air resistance (in the course of cushioning) and other minor ingredients. To guarantee safety, in the course of falling, the boxes can absorb at least as much as the formerly related decrement of energy.The Most Desirable Box - An Optimal ModelThe most desired box is such that the effect of cushion can be obtained with fewer boxes and lower height. The more energy the box can absorb per unit volume, the fewer boxes we need to cushion the fall, and it is more convenient to avoid being seen by camera. This means a most desirable box is one with largest .According to the assumptions, to guarantee safety, the force that the stunt person can endure should be no more than the maximum limit. HenceF M (4g+g) = 5 M g To satisfy this constraint, we only need to be sure that 5 Mg ,where = P =.P is the maximum compression strength of a box and shows the number of boxes cushioning the fall simultaneously, which is determined by l and d of a box. So the constraint is modified to 5 Mg To solve this problem we propose an optimal Model. Object: Max= Subject to: 5 Mg Using the Searching Algorithm with Turbo C (see Appendix B), we get the sides of the most desirable box:L = 0.48m, d = 0.30m , and h = 0.20m .Cushioning the FallWe define Wva = and Wha = ,where n denotes the number of boxes to cushion the fall.To guarantee safety, the following constraints must be satisfied: where S the horizontal displacement of MASP between reaching the box and the ground. Since the cushion lasts only for a very short time, the horizontal motion can be viewed approximately as uniform speed-reduction (whose acceleration is a constant). This system of equations can be modified to:where nh denotes the number of boxes to cushion the fall horizontally.The resultsThe size of the boxes is the same as that of the most desired box.The altitude of the top point is 4.5m ( hm =4.5m). Using Turbo C, we know that 270 boxes should be used if they are piled into a 6-storeyed rectangular solid.Considering that MASP is going down constantly, we can stack the boxes in the form of a step. The number of boxes is reduced to 35+37+39+311+313+315=180.The Test of ModelGiven the mass of MASP, we calculate the size of a box and the number needed if theyre stacked into a rectangular solid. Then we calculate the number of boxes. We adopt different papers, resulting in different Pm. The relationship between the number of needed boxes and Pm is shown as Figure 4: Figure 4.We fix hmax=4.5m and Pm=2000N/m, and get: Figure 5.We fix M=2000Kg and Pm=2000N/m , and get:Figure 6.We analyse the effects of modification to the boxes and different combined weights and different jump heights. The results are as follows:TableHmax(m) M(kg) NumberV(m3) Length(m)Width(m)Heigth(m)4.0 150 2303.7230.48 0.16 0.21 170 4665.9520.48 0.14 0.19 190 3343.8490.48 0.12 0.20 210 5255.9980.48 0.14 0.17 230 7676.2630.48 0.10 0.17 250 10288.8830.48 0.12 0.155.5 150 706 11.400.48 0.16 0.21 170 119315.230.48 0.14 0.19 190 102311.780.48 0.12 0.20 210 134415.350.48 0.14 0.17 230 195616.030.48 0.10 0.17 250 231319.980.48 0.12 0.157.0 150 144223.260.48 0.16 0.21 170 225628.810.48 0.14 0.19 190 208824.050.48 0.12 0.20 210 3024 34.550.48 0.14 0.17 230 371530.310.48 0.10 0.17 250 482641.700.48 0.12 0.158.5 150 243839.320.48 0.16 0.21 170 315140.240.48 0.14 0.19 190 352940.650.48 0.12 0.20 210 472653.990.48 0.14 0.17 230 601849.100.48 0.10 0.17 250 825371.310.48 0.12 0.1510 150 324652.350.48 0.16 0.21 170 477460.950.48 0.14 0.19 190 534561.580.48 0.12 0.20 210 680577.740.48 0.14 0.17 230 887372.410.48 0.10 0.17 2501142498.700.48 0.12 0.15Strengths and Weaknesses Considering a single cardboard box first, an optimal size of a box can be worked out. The size of a box remains the same only with the weight changeable, so that the calculation can be simplified. The size of a box is obtained by a part optimization, not necessarily the most favorable as a whole, therefore, the possible boxes may be increased.Appendix AThe track is shown as Figure 1and 2.Vx = ax = To solving the system of equations ,we can only turn help to computer for solution using numerical approximation. Thus we do the following analysis assuming that the force F is the linear function of velocity ( F = k v ).Based on this assumption , the horizontal velocity and acceleration are:Vx = ax = The equations describing the motion of MASP arewhere x the horizontal distance between MASP and point D.From these, we get x(t) =Since both k and t are small enough (k is no more than 0.1), x(t) = y(t) =The horizontal velocity is vx= and the acceleration is ax=, which shows that the motorcycle reach the top point at tc =.The time required to move from point A to point C and from point C to point B is the same, which we defined t. The motorcycle reaches point A and point B at ta=tc-t and tb=tc+t, whose horizontal velocities are Va =Vb =Then 2t= t= = The altitude at point A and point B is:HA=HB=.To jump over the elephant, HA and HB must be no smaller than h. Given an , we get the range of velocity and the relationship of and Vmin .The vertical velocity is Vy=.So the velocity of motorcycle is V=.If hm denotes the the total height of the cushioning boxes, and tm denotes the time the motorcycle reaches this altitude, then: hm= t= Substituting t into V=, we get: V = which is useful for determining the height of boxes.Appendix BThe C Programme#include math.hmain() int j; float c1=6.459,c2=8.0798,max=0.0,w; float a,b,m,l,d,h,pm,q,f,n,n1,n2,hm,i5=0,0,0,0,0; clrscr(); for(hm=4;hm=10;hm+=1,max=0) for(m=150;m=250;m=m+10) for(pm=10;pm=40;pm=pm