389 lines
12 KiB
Python
389 lines
12 KiB
Python
"""
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2026/3/29
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"""
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import struct
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import math
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def calc_telem(value: float, method: str, param_str: str, ref_cache: dict) -> float:
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#def telem_calc(value: float, method: str, param_str: str, ref_cache: dict) -> float:
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try:
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x = value
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# 空方法：原值
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if not method:
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return x
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# 数值拟合 y = a*x + b
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if method == "数值拟合":
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a, b = map(float, param_str.split(','))
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return a * x + b
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# 电压1：V = D/255*5.1 D无符号整型
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elif method == "电压1":
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D = int(x)
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return D / 255.0 * 5.1
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# 电压6：V = D*20/65535 D有符号整型
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elif method == "电压6":
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D = int(x)
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return D * 20.0 / 65535.0
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# 电压3：V = D*0.000305185 D有符号整型
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elif method == "电压3":
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D = int(x)
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return D * 0.000305185
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# 电压2：V= D *10/4096-5 D有符号整型
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elif method == "电压2":
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D = int(x )
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return D * 10 / 4096 - 5
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# # 电阻拟合1
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# elif method == "电阻拟合1":
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# D = int(x)
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# V = D / 255.0 * 5.1
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# ref_key = param_str.strip()
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# Vref = ref_cache.get(ref_key, 5.1)
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# if abs(Vref - V) < 1e-12:
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# return 0.0
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# Rr = 10000.0
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# return V * Rr / (Vref - V)
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# 电阻拟合1
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elif method == "电阻拟合1":
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D = int(x)
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V = D / 255.0 * 5.1
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# V基准固定为 5.0V
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Vref = 5.0
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if abs(Vref - V) < 1e-12:
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return 0.0
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Rr = 10000.0
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return V * Rr / (Vref - V)
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# # 特殊公式8（按你描述完整实现）
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# elif method == "特殊公式8":
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# raw = int(x)
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# if param_str:
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# parts = param_str.split(',')
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# else:
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# parts = []
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# flag = int(parts[0]) if len(parts) >= 1 else 0
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# if flag == 0:
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# # 遥测状态字：16位处理
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# w = raw & 0xFFFF
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# bh = (w >> 8) & 0xFF
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# bl = w & 0xFF
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# bh8 = bh + 0x8
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# bl8 = bl + 0x8
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# h4 = (bh8 >> 4) & 0xF
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# l4 = (bl8 >> 4) & 0xF
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# B1 = (l4 << 4) | h4
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# else:
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# # 遥控状态字：8位直接用
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# B1 = raw & 0xFF
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# # 后面位比较逻辑你需要我也可以补全
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# return float(B1)
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elif method == "特殊公式8":
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"""
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最新规则：
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1) 原始值取整
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2) 低 16bit：低位
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3) 高 4bit ：高位
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4) 合并 = (高位4bit << 16) | 低位16bit
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5) 结果 = 合并值 × 0.00205479
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"""
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raw_int = int(x) & 0xFFFFFFFF # 取32位无符号
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# 低位 16bit
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low16 = raw_int & 0xFFFF
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# 高位 4bit（最后4bit）
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high4 = (raw_int >> 28) & 0x0F
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# 合并 20bit：高4位 <<16 | 低16位
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combined = (high4 << 16) | low16
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# 乘以系数
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result = combined * 0.00205479
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return float(result)
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# 状态量 / 10进制：不处理
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elif method in ["状态量", "10进制"]:
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return x
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return x
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except Exception as e:
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return value
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import struct
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import math
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def poly4_calc(D):
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"""正向 4 次多项式计算 y"""
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y = (-49.736 * (D ** 4)
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+ 383.38 * (D ** 3)
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- 1068.5 * (D ** 2)
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+ 1309.9 * D
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- 684.71)
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return y
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def inverse_poly4_fit(y):
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"""
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逆运算：已知 y → 求 D
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适用于：y ∈ [-112, 42]
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"""
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if not (-112 <= y <= 42):
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return 0.0 # 超范围保护
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# 二分法区间（根据你的遥测范围实测得出）
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left = 1.5
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right = 4.5
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eps = 1e-5
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for _ in range(50):
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mid = (left + right) / 2
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current_y = poly4_calc(mid)
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if current_y < y:
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left = mid
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else:
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right = mid
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return (left + right) / 2
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# 固定系数
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Rr = 10000.0
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A = -6.01188
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B = 4622.53337
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C = -86421.72414
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def calc_t_from_V(V):
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"""正向：V → t（内部用）"""
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if V >= 5.0:
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V = 4.9999
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if V <= 0:
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V = 0.0001
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R = V * Rr / (5.0 - V)
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lnR = math.log(R)
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sqrt_val = math.sqrt(B**2 - 4 * C * (A - lnR))
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t = 2 * C / (sqrt_val - B) - 273.15
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return t
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def calc_V_from_t(t_target):
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"""
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逆运算：已知温度 t → 求遥测电压 V
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:param t_target: 目标温度 ℃
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:return: V (V)，精度 0.0001
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"""
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# 电压范围 0 ~ 5V
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left = 0.0001
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right = 4.9999
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eps = 1e-5
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# 二分法迭代 50 次足够精确
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for _ in range(50):
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mid = (left + right) / 2
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t_mid = calc_t_from_V(mid)
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if t_mid < t_target:
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left = mid
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else:
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right = mid
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return (left + right) / 2
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def inverse_quadratic_fit(y, coeffs, y_min, y_max):
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"""
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通用二次多项式逆运算（3个拟合参数）
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公式：y = a2*D² + a1*D + a0
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已知 y → 求 D（只返回合法正根）
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参数：
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y: 目标物理值
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coeffs: 拟合系数 [a2, a1, a0]
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y_min: y 有效最小值
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y_max: y 有效最大值
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返回：
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D: 原始码值
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"""
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# 1. 边界限幅
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y_clamped = max(y_min, min(y, y_max))
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# 2. 取出系数
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a2, a1, a0 = coeffs
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# 3. 构建一元二次方程标准形式：A*D² + B*D + C = 0
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A = a2
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B = a1
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C = a0 - y_clamped
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# 4. 求判别式
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delta = B ** 2 - 4 * A * C
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if delta < 0:
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delta = 0 # 防止无实根
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# 5. 求正根（遥测D一定为正）
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D = (-B + (delta ** 0.5)) / (2 * A)
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# 6. 保证 D 不为负
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return max(D, 0.0)
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def calc_telem_reverse(result: float, method: str, param_str: str) -> float:
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"""
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【逆运算】
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输入：A 软件发来的 结果值
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输出：反推出来的 原始值 D
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完全按你提供的处理方法反向计算
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"""
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try:
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y = result
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# ===================== 空方法：直接返回 =====================
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if not method:
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return y
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# ===================== 数值拟合：Y = a*x + b → x=(y-b)/a =====================
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if method == "数值拟合":
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a, b = map(float, param_str.split(','))
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x = (y - b) / a
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return x
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# ===================== 电压1：V = D/255*5.1 → D=V*255/5.1 =====================
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elif method == "电压1":
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D = y * 255.0 / 5.1
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return float(int(round(D))) # 原始是无符号整型
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# ===================== 电压6：V = D*20/65535 → D=V*65535/20 =====================
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elif method == "电压6":
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D = y * 65535.0 / 20.0
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return float(int(round(D))) # 有符号整型
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# ===================== 电压3：V = D*0.000305185 → D=V / 0.000305185 =====================
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elif method == "电压3":
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D = y / 0.000305185
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return float(int(round(D))) # 有符号整型
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# ===================== 电压2：V电压2 = D *10/4096-5，其中D为无符号整型，计算时转化成十进制数。 → D=（V + 5）*409.6 =====================
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elif method == "电压2":
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D = (y + 5) * 409.6
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return float(int(round(D))) # 有符号整型
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# ===================== 电阻拟合1 → 逆运算 =====================
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# 公式：
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# V = D/255*5.1
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# R = V*10000/(5.0 - V)
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# 逆推 D
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#2026/3/30手算验证公式是正确的
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elif method == "电阻拟合1":
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R = y
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Vref = 5.0
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Rr = 10000.0
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# 逆推 V
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if R <= 0.0:
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V = 0.0
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else:
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V = (R * Vref) / (R + Rr)
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# 逆推 D
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D = V * 255.0 / 5.1
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return float(int(round(D)))
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# ===================== 特殊公式8（逆运算） =====================
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# 正向：combined = (high4 <<16) | low16 result = combined * 0.00205479
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# 逆向：combined = y / 0.00205479
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# high4 = (combined >> 16) & 0xF
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# low16 = combined & 0xFFFF
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# raw = (high4 << 28) | low16
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elif method == "特殊公式8":
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combined = int(round(y / 0.00205479))
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low16 = combined & 0xFFFF
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high4 = (combined >> 16) & 0x0F
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raw = (high4 << 28) | low16
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return float(raw)
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# ===================== 状态量 / 10进制：直接返回 =====================
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elif method in ["状态量", "10进制"]:
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return y
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elif method == "特殊公式11-1":
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"""
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M057JG023 / M057JG024 专用逆运算
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输入：物理值 y
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输出：原始值 m（用于组帧）
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"""
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# if y < (0.0973 * 1840 + 180): # 区间1：y < 359.032
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# m = (y - 180) / 0.0973
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# elif y > (0.0974 * 1840 - 179.46): # 区间2：y > -0.444
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# m = (y + 179.46) / 0.0974
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# else: # m=1840
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# m = 1840.0
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#2026/5/7修正公式
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if y < (180): # 区间1：
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m = (y + 179.46) / 0.0974
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elif y > (180): # 区间2：
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m = (y - 180) / 0.0973
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else: # m=1840
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m = 0.0
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# 组帧需要整数，返回四舍五入
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return float(round(m))
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elif method == "电压2，数值拟合":
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"""
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逆运算，先反解数值拟合，再反推电压2
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"""
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a, b = map(float, param_str.split(','))
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x = (y - b) / a
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D = (x + 5) * 409.6## 电压2
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return float(int(round(D))) # 有符号整型
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elif method == "电压1，数值拟合":
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"""
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逆运算，先反解数值拟合，再反推电压2
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"""
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a, b = map(float, param_str.split(','))
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x = (y - b) / a
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D = x * 255.0 / 5.1## 电压1
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return float(int(round(D))) # 有符号整型
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elif method == "电压1，数值拟合5":
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"""
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逆运算，先反解数值拟合（5次方程），再反推电压2
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"""
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x = inverse_poly4_fit(y) ##2分法求解5次方程
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D = x * 255.0 / 5.1## 电压1
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return float(int(round(D))) # 有符号整型
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elif method == "电阻拟合1,MF501(电阻)":
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x = calc_V_from_t(y) #2分法
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R = x
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Vref = 5.0
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Rr = 10000.0
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# 逆推 V
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if R <= 0.0:
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V = 0.0
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else:
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V = (R * Vref) / (R + Rr)
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# 逆推 D
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D = V * 255.0 / 5.1
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return float(int(round(D)))
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elif method == "电压1，数值拟合3":
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a, b,c,y_min,y_max = map(float, param_str.split(','))
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coeffs = [a,b,c]
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return inverse_quadratic_fit(y, coeffs, y_min, y_max)
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elif method == "电压6，数值拟合":
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"""
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逆运算，先反解数值拟合，再反推电压2
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"""
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a, b = map(float, param_str.split(','))
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x = (y - b) / a
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D = x * 65535.0 / 20.0 #电压6
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return float(int(round(D))) # 有符号整型
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except Exception as e:
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return result |