由于两个柏拉图式实体的问题,我在发表我的论文时遇到了麻烦,我正在为它们的SA:V编码:立方体和八面体
通常,当我们推导出立方体和八面体的SA:V比时,它将分别得出6/a和(3*sqrt(6))/a
当你把这些比率作为一个比率,你会得到一个恒定的非1:1的比率,对于所有大小的区域,那么,我的输出是一个1:1的关系吗
输出(单击链接):
用户请求、制表和绘图的所有实例中两种形状的相关代码(忽略二十面体代码):第一个实例:
elif name_of_polyhedron == 'Octahedron':
a = d/math.sqrt(2)
SA_vex_octa = ((2*math.sqrt(3))*a**2)
V_vex_octa = (((math.sqrt(2))/(3))*a**3)
ratio_vex_octa = (SA_vex_octa/V_vex_octa)
print('The surface area of your octahedron is as follows:' +str(SA_vex_octa)+ 'm^2')
print('The volume of your octahedron is as follows:' +str(V_vex_octa)+ 'm^3')
print('The surface area to volume ratio of your octahedron is as follows:'
+str(ratio_vex_octa))
print('See how your ratio compares below!')
elif name_of_polyhedron == 'Icosahedron':
a = 4*(1/(math.sqrt(10+2*math.sqrt(5))))*((1/2)*d)
SA_vex_icosa = 5*(a**2)*math.sqrt(3)
V_vex_icosa = (5/12)*(a**3)*(3+sqrt(5))
ratio_vex_icosa = SA_vex_icosa/V_vex_icosa
print('The surface area of your icosahedron is as follows:' +str(SA_vex_icosa)+ 'm^2')
print('The volume of your icosahedron is as follows:' +str(V_vex_icosa)+ 'm^3')
print('The surface area to volume ratio of your icosahedron is as follows:'
+str(ratio_vex_icosa))
print('See how your ratio compares below!')
elif name_of_polyhedron == 'Cube':
a = d/math.sqrt(3)
SA_vex_cube = 6*a**2
V_vex_cube = a**3
ratio_vex_cube = SA_vex_cube/V_vex_cube
print('The surface area of your cube is as follows:' +str(SA_vex_cube)+ 'm^2')
print('The volume of your cube is as follows:' +str(V_vex_cube)+ 'm^3')
print('The surface area to volume ratio of your cube is as follows:' +str(ratio_vex_cube))
print('See how your ratio compares below!')
#Second Instance
a_4 = d/math.sqrt(2)
SA_vex_octa = (2*math.sqrt(3))*(a_4)**2
V_vex_octa = ((math.sqrt(2))/(3))*(a_4)**3
ratio_vex_octa = SA_vex_octa/V_vex_octa
a_5 = 4*(1/(math.sqrt(10+2*math.sqrt(5))))*((1/2)*d)
SA_vex_icosa = 5*((a_5)**2)*math.sqrt(3)
V_vex_icosa = (5/12)*((a_5)**3)*(3+math.sqrt(5))
ratio_vex_icosa = SA_vex_icosa/V_vex_icosa
a_6 = d/math.sqrt(3)
SA_vex_cube = 6*(a_6)**2
V_vex_cube = (a_6)**3
ratio_vex_cube = SA_vex_cube/V_vex_cube
#Third Instance
a_3 = (2/math.sqrt(6))*d
a_4 = d/math.sqrt(2)
a_5 = (2/(math.sqrt(10+2*math.sqrt(5))))*d
a_6 = d/math.sqrt(3)
a_7 = (2/(math.sqrt(3)*(1+math.sqrt(5)))*d)
a_8 = ((2/(math.sqrt(10+2*math.sqrt(5))))*d)
a_9 = ((2/(math.sqrt(50+22*math.sqrt(5))))*d)
a_10 = ((3/(math.sqrt(3)*(3+math.sqrt(5))))*d)
a_11 = ((2/(math.sqrt(50+22*math.sqrt(5))))*d)
SA_vex_tetra = (((a_3)**2)*math.sqrt(3))
V_vex_tetra = ((((a_3)**3)/12)*math.sqrt(2))
ratio_vex_tetra = [SA_vex_tetra/V_vex_tetra]
SA_vex_octa = (2*math.sqrt(3))*(a_4)**2
V_vex_octa = ((math.sqrt(2))/(3))*(a_4)**3
ratio_vex_octa = SA_vex_octa/V_vex_octa
SA_vex_icosa = 5*((a_5)**2)*math.sqrt(3)
V_vex_icosa = (5/12)*(a_5)**3*(3+math.sqrt(5))
ratio_vex_icosa = SA_vex_icosa/V_vex_icosa
SA_vex_cube = 6*(a_6)**2
V_vex_cube = (a_6)**3
ratio_vex_cube = SA_vex_cube/V_vex_cube
它们出现的表格/图表:
import matplotlib.pyplot as plt
from tabulate import tabulate
z = [('Tetrahedon',a_2,'Platonic',d,SA_vex_tetra,V_vex_tetra,ratio_vex_tetra),
('Octahedron',a_2,'Platonic',d,SA_vex_octa,V_vex_octa,ratio_vex_octa),
('Icosahedron',a_2,'Platonic',d,SA_vex_icosa,V_vex_icosa,ratio_vex_icosa),
('Cube',a_2,'Platonic',d,SA_vex_cube,V_vex_cube,ratio_vex_cube),
('Dodecahedron',a_2,'Platonic',d,SA_vex_dodeca,V_vex_dodeca,ratio_vex_dodeca),
('Cuboctahedron',a_2,'Archimedes',d,SA_vex_cubocta,V_vex_cubocta,ratio_vex_cubocta),
('Rhombicuboctahedron',a_2,'Archimedes',d,SA_vex_rhocubocta,V_vex_rhocubocta,ratio_vex_X),
('Snub Cube',a_2,'Archimedes',d,SA_vex_scube,V_vex_scube,ratio_vex_scube),
('Snub Dodecahedron',a_2,'Archimedes',d,SA_vex_sndodeca,V_vex_sndodeca,ratio_vex_sndodeca),
('Rhombicosidodecahedron',a_2,'Archimedes',d,SA_vex_ridodeca,V_vex_ridodeca,ratio_vex_ridodeca),
('Truncated Octahedron',a_2,'Archimedes',d,SA_vex_ridodeca,V_vex_ridodeca,ratio_vex_sndodeca),
('Deltoidal Icositetrahedron','Catalan',a_2,d,SA_vex_delicotetra,V_vex_delicotetra,ratio_vex_X),
('Great Dodecahedron',a_1,'Kepler-Poinsot',d,SA_cav_gdodeca,V_cav_gdodeca,ratio_cav_gdodeca),
('Great Icosahedron',a_1,'Kepler-Poinsot',d,SA_cav_gicosa,V_cav_gicosa,ratio_cav_gicosa),
('Great-Stellated Dodecahedron',a_1,'Kepler-Poinsot',d,SA_cav_gsdodeca,V_cav_gsdodeca,ratio_X_X),
('Small-Stellated Dodecahedron',a_1,'Kepler Poinsot',d,SA_cav_ssdodeca,V_cav_ssdodeca,X),
('Stellated Octahedron',a_1,'Da Vinci',d,SA_cav_stocta,V_cav_stocta,ratio_cav_stocta),
('Medial Rhombic Triacontahedron',a_1,'A-R',(Wenninger)',d,SA_cav_mrtria,V_cav_mrtria,X),
('Dodecadodecahedron',a_1,'A-R (Wenninger)',d,SA_cav_ddodeca,V_cav_ddodeca,ratio_cav_ddodeca),
('Medial Triambic Icosahedron',a_1,'A-R', (Wenninger)',d,SA_cav_mticosa,V_cav_mticosa,X),
('Small Ditrigonal Icosidodecahedron',a_1,'A-R',(Wenninger)',d,SA_cav_sdicosi,V_cav_sdicosi,X),
('Excavated Dodecahedron',a_1,'A-R',(Wenninger)',d,SA_cav_exdodeca,V_cav_exdodeca,X),
('Sphere',a_12,a_12,d,SA_sphere,V_sphere,ratio_sphere)]
table_1 = tabulate(z, headers=['Shape','Type','Subtype','C.D. (m)',
'SA (m^2)','V(m^3)','SA:V'], tablefmt='fancy_grid') #orgtbl or f or pretty
print(table_1)
Platonic_Array_Ratios = [ratio_vex_tetra,ratio_vex_octa,
ratio_vex_icosa,ratio_vex_cube,
ratio_vex_dodeca,ratio_sphere]
plt.title('Surface Area to Volume Ratio of Platonic Polyhedra Against Referential Sphere for
Given
Diameter')
plt.barh(['Tetrahedron', 'Octahedron',
'Icosahedron', 'Cube',
'Dodecahedron', 'Sphere'], Platonic_Array_Ratios)
plt.show()
# Platonic
plt.scatter(d,ratio_vex_tetra,label='Tetrahedron', color='b')
plt.scatter(d,ratio_vex_octa,label='Octahedron', color='g')
plt.scatter(d,ratio_vex_icosa,label='Icosahedron', color='y')
plt.scatter(d,ratio_vex_cube,label='Cube', color='m')
plt.scatter(d,ratio_vex_dodeca,label='Dodecahedron', color='c')
plt.scatter(d,ratio_sphere,label='Referential Sphere', color='r')
plt.axvline(x=d_vertical_slash, color='k')
plt.title('Comparison of SA:V of Platonic polyhedra against Referential Sphere with D(m)')
plt.xlabel('Circumspherical Diameter/Diagonal (m)')
plt.ylabel('Ratio Index')
plt.grid(alpha=.4,linestyle='--')
plt.xscale("log")
plt.legend(loc = 1)
plt.show()
我已经仔细地用手工和在线小部件检查了四倍的计算结果……它们都是正确的,并且在程序输出的所有计算中都相互匹配,但仍然有一些错误。似乎一个正在覆盖另一个,但在哪里
注意:此代码中没有错误消息。我也为代码太长而道歉。在某些情况下,为了再现性或只是为了提供更大的上下文,我不得不将所有内容复制并粘贴在一起。X只是一个用于格式化的占位符
您正在计算“给定直径的柏拉图多面体相对于参考球体的表面积体积比”[黑体添加]
因此,在比较立方体和八面体之前,先比较立方体和球体的直径,以及八面体和球体的直径
(我想这意味着外接球;我不打算检查内接球。)
球体的SA/V比为6/d,其中d为直径
边长为a的立方体的最大直径为sqrt(3)a,因此立方体/球体比率为(6/a)/(6/(sqrt(3)a))=sqrt(3)
边长为a的八面体的最大直径为sqrt(2)a,因此八面体/球体比率为(3sqrt(6)/a)/(6/(sqrt(2)a))=sqrt(3)
由于与球体的比较,比率为1:1
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