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functions_backup.py
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131 lines (118 loc) · 5.1 KB
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#coding:utf-8
def calLine(points):
#计算二维直线参数
#传入两个点坐标的数组[x_min,z_max,x_max,z_min] 返回直线的两个参数
#使用 np.linalg.solve 计算
import numpy as np
p = points
a = np.array([[p[0],1],[p[2],1]])
b = np.array([p[1],p[3]])
params = np.linalg.solve(a,b)
return params
def getConvertMatrix(points):
#计算转换矩阵 以筛网建立坐标系
#传入筛网的两个点
#转换矩阵 新坐标系基底在旧坐标系里的投影坐标
#转换矩阵 [cos(a),sin(a);-sin(a),cos(a)]
p = points
dp = np.array([abs(p[0]-p[2]),abs(p[1]-p[3])])
modp = np.sqrt(sum(dp**2)) #求向量的模
sinp = dp[1]/modp
cosp = dp[0]/modp
con_matrix = np.matrix([[cosp,sinp],[-sinp,cosp]])
return con_matrix
def getEndTime(ptc_tim,scn_tim,tim_ls):
#计算筛分结束时刻 1、以筛面上颗粒数量减少率或最大颗粒数量的10%来定；2、以筛分效率稳定时刻来定
###########方式1：###########
import numpy as np
tim_len = len(ptc_tim)
num_ptc = np.ones(tim_len)
for ti in range(tim_len):
#统计某一时刻 筛面上的颗粒数量 num_ptc
# ti = 0 #指定时刻
kd = calLine(scn_tim[ti])
ptc_xz = ptc_tim[ti][:,0:2]
bool_up = (ptc_xz[:,1]-(kd[0]*ptc_xz[:,0]+kd[1]))>0
num_ptc[ti] = sum(bool_up) #得出筛面上颗粒总数
id_edt = sum(num_ptc>max(num_ptc)*0.02) #以最后一个大于最大颗粒数2%的时刻为结束时刻
print('there still has {} on the deck when time={}'.format(num_ptc[id_edt-1],tim_ls[id_edt-1]))
#注意！！要想取出对应时刻，需要用 times(id_edt-1)
# import matplotlib.pyplot as plt
# plt.plot(np.arange(tim_len),num_ptc) #画出该实验下 各个时刻时筛面上颗粒数量变化
# plt.show()
########方式2：#########
# #筛分效率的变化趋势 判断筛分截止时刻
# tim_len = len(tim_ls)
# eff = np.ones(tim_len)
# for i in range(tim_len):
# eff[i] = calScrEff(ptc_tim[i],scn_tim[i])
# # plt.plot(tim_ls,eff);plt.show()
# id_ed_eff = sum(eff < max(eff)*0.98)
return id_edt-1
def calScrEff(ptc,scn_tim):
#计算单位时间筛分效率 传入一个时刻下的 颗粒数据 和 筛网数据
dim = 0.9 #指定分类粒径
z_min = -44
x_max = scn_tim[2]
kb = calLine(scn_tim)
bool_und = (ptc[:,0]<x_max)&(ptc[:,1]>z_min)&((ptc[:,1]-ptc[:,0]*kb[0]-kb[1])<0)
ptc_und = ptc[bool_und]
dns = 2678 #密度2678千克每立方米
dim = dim/1000 #将 毫米 单位化成 米
mass_std = ((np.pi/6)*dns*dim**3)*1000 #把质量的kg单位变成g单位 分离粒径颗粒质量
bool_all_sml = ptc[:,2]<mass_std
bool_und_sml = ptc_und[:,2]<mass_std
eff_sml = sum(ptc_und[bool_und_sml,2])/sum(ptc[bool_all_sml,2])
eff_lag = sum(ptc_und[~bool_und_sml,2])/sum(ptc[~bool_all_sml,2])
eff = eff_sml - eff_lag
# plt.plot(ptc[bool_all_sml,0],ptc[bool_all_sml,1],'.',c='k',alpha=0.3) #画小颗粒分布
# plt.plot(ptc[~bool_all_sml,0],ptc[~bool_all_sml,1],'.',c='g',alpha=0.1) #画大颗粒分布
# plt.show() #验证图
# return eff,eff_sml,eff_lag
return eff
def stabTim(ptc_tim,scn_tim):
########如何返回稳定筛分 时间段########
#获得入料柱 直接设定 x=-66 看之后其他实验是否适合再改 or x_max z轴最大的1%的颗粒的x最大值？？
# plt.scatter(ptc_tim[0][:,0],ptc_tim[0][:,1],marker='.');plt.show()
#筛上颗粒 area1:x1>-66 x1<-56; area2:x2>scn_x_max-10 x2<scn_x_max;
#在这两个区域内分别有(0.015)筛上颗粒的颗粒数量 根据后续实验 可能 要改变0.015取值
import numpy as np
x_bod = -66
tim_len = len(ptc_tim)
are1 = np.empty(tim_len);are2 = np.empty(tim_len);total = np.empty(tim_len)
for ti in range(tim_len):
ptc = ptc_tim[ti]
kb = calLine(scn_tim[ti])
bool_up = (ptc[:,0]>x_bod)&(ptc[:,0]<scn_tim[ti,2])&((ptc[:,1]-ptc[:,0]*kb[0]-kb[1])>0)
num_ptc_up = sum(bool_up)
ptc_up = ptc[bool_up]
num_ptc_are1 = sum((ptc[:,0]>x_bod)&(ptc[:,0]<x_bod+10)&bool_up)
num_ptc_are2 = sum((ptc[:,0]>scn_tim[ti,2]-10)&(ptc[:,0]<scn_tim[ti,2])&bool_up)
are1[ti] = num_ptc_are1
are2[ti] = num_ptc_are2
total[ti] = num_ptc_up
bool_stab = (are1>max(total)*0.015)&(are2>max(total)*0.015) # 布尔矩阵，可用于稳定筛分阶段时间的取值
id_tim_stab = np.arange(tim_len)[bool_stab] #取出稳定筛分阶段的 时刻 的标号
return id_tim_stab[0],id_tim_stab[-1] #只返回 起始 和 截止 时刻标号
def calPorosity():
#计算松散性
pass
def calStratification():
#计算分层状态
pass
def calCollision():
#计算触筛概率
pass
def calPenetrating():
#计算透筛概率
pass
def timEff98(tim_ls,ptc_tim,scn_tim):
#计算达到最大筛分效率98%的时刻 表征一种快慢 有没有用？？
tim_len = len(tim_ls)
eff = np.ones(tim_len)
for i in range(tim_len):
eff[i] = calScrEff(ptc_tim[i],scn_tim[i])
# plt.plot(tim_ls,eff);plt.show()
id_eff = sum(eff < max(eff)*0.98)
tim = tim_ls[id_eff]
return tim