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\begin{align*} (\frac{d_m}{2})^2-(\frac{d_n}{2})^2...

Creato il: 6 maggio 2025

Risposto usando O3 da Chat01

Domanda

\begin{align*}
(\frac{d_m}{2})^2-(\frac{d_n}{2})^2&=(m - n)r\lambda
\frac{d_m^2 - d_n^2}{4}&=(m - n)r\lambda
r&=\frac{d_m^2 - d_n^2}{4(m - n)\lambda}
\end{align*}

\lambda\lambda = 589.3nm
仪器不确定度0.004mm计算R平均值和不确定度

dm2d_m^2 dm2−dm−52d_m^2 - d_{m-5}^2 mm2^2

/mmd/mmddmd_m/mmdm2d_m^2/mm2^2dm2−dm−52d_m^2 - d_{m-5}^2/mm2^2
2025.08019.4485.63231.7198.679
1925.01819.5305.48830.1189.270
1824.89119.6145.27727.8478.531
1724.80519.7005.10526.0618.253
1624.72719.7784.94924.4938.204
1524.62019.8204.80023.040
1424.52419.9584.56620.848
1324.43520.0404.39519.316
1224.37020.1504.22017.808
1124.27020.2344.03616.289

Python

python
import numpy as np, math
deltas = np.array([8.679, 9.270, 8.531, 8.253, 8.204]) # mm^2
lam = 5.893e-4 # mm
C = 20*lam
R = deltas / C
R

array([736.38214831, 786.52638724, 723.82487697, 700.23757   ,
       696.08009503])

Python

python
R_mean = R.mean()
R_std = R.std(ddof=1) # sample std
R_mean, R_std

(728.6102155099271, 36.40330034505399)

Python

python
d_m = np.array([5.632, 5.488, 5.277, 5.105, 4.949])
d_n = np.array([4.800, 4.566, 4.395, 4.220, 4.036]) # d_{m-5}
deltaR_instr = []
for dm, dn in zip(d_m, d_n):
    dDelta = math.sqrt((2*dm*0.004)**2 + (2*dn*0.004)**2)
    deltaR = dDelta / C
    deltaR_instr.append(deltaR)
deltaR_instr

[5.0228821184311805,
 4.8458052429069935,
 4.661472525036131,
 4.49577451671188,
 4.334682195521749]

Python

python
N = len(R)
u_repeat = R_std / math.sqrt(N)
u_instr = np.mean(deltaR_instr) # or worst-case
combined = math.sqrt(u_repeat**2 + u_instr**2)
(R_mean, u_repeat, u_instr, combined)

(728.6102155099271, 16.280050835376453, 4.672123319721587, 16.937201407467757)

Pensiero

Final uncertainty calculation

I’ve combined the statistical uncertainty with the instrument’s systematic uncertainty. The final uncertainty comes to about 16.9 mm, and maybe it’s better to present this as R = 729 ± 17 mm (using the maximum instrument uncertainty of 5.0 mm). I’ll create an explanation in both Chinese and English, showing intermediate steps and the final result. I’m thinking about using a table for clarity, but I’ll skip the detailed code for the user.

Thought for 1m 4s

Risposta

1. 取 m 与 m − 5 两个干涉环差法的公式

R=dm2−dm−524(m−(m−5))λ=dm2−dm−5220λ,λ=589.3  nm=5.893×10−4  mmR=\frac{d_m^{2}-d_{m-5}^{2}}{4(m-(m-5))\lambda} =\frac{d_m^{2}-d_{m-5}^{2}}{20\lambda},\qquad \lambda = 589.3\;\text{nm}=5.893\times10^{-4}\;\text{mm}

常数
C=20λ=0.011 786  mmC=20\lambda = 0.011\,786\;\text{mm}.

2. 逐列计算

mdm2−dm−52d_m^2-d_{m-5}^2 / mm^2Ri=(dm2−dm−52)/CR_i=(d_m^2-d_{m-5}^2)/C / mm
208.679736.4
199.270786.5
188.531723.8
178.253700.2
168.204696.1

3. 平均值与重复性不确定度

Rˉ=728.6  mm,s=36.4  mm,  urep=sN=16.3  mm.\bar R = 728.6\;\text{mm},\qquad s = 36.4\;\text{mm},\; u_\text{rep}=\frac{s}{\sqrt{N}}=16.3\;\text{mm}.

4. 仪器读数不确定度的传播（每条）

单次直径读数的不确定度：
σd=±0.004  mm\sigma_d = \pm0.004\;\text{mm}.

uRi,inst=(2dmσd)2+(2dm−5σd)2C≈4.3–5.0  mm.u_{R_i,\text{inst}} =\frac{\sqrt{(2d_m\sigma_d)^2 + (2d_{m-5}\sigma_d)^2}}{C} \approx 4.3\text{–}5.0\;\text{mm}.

采用其平均值 uinst=4.7  mmu_\text{inst}=4.7\;\text{mm}.

5. 合成标准不确定度

uc=urep2+uinst2=16.32+4.72  mm≈16.9  mm.u_c = \sqrt{u_\text{rep}^2 + u_\text{inst}^2} = \sqrt{16.3^2 + 4.7^2}\;\text{mm} \approx 16.9\;\text{mm}.

结果

R=(7.29±0.17)×102 mm  (k=1,  uc=17 mm)=72.9±1.7 cm.\boxed{R = (7.29 \pm 0.17)\times10^{2}\ \text{mm}} \; \bigl(k=1,\;u_c=17\ \text{mm}\bigr) = 72.9 \pm 1.7\ \text{cm}.
  • 平均半径 R ≈ 72.9 cm
  • 合成标准不确定度 ≈ 1.7 cm (取展伸系数 k = 1)。

Domanda

不确定度使用重复性不确定度和仪器不确定度平方和的根号
2.生成一个docx文件

Python

python
# Data
data = [
    (20, 5.632, 4.800),
    (19, 5.488, 4.566),
    (18, 5.277, 4.395),
    (17, 5.105, 4.22),
    (16, 4.949, 4.036)
]
σ_d = 0.004
λ = 5.893e-4
C = 20 * λ  # mm
results = []
for m, d_m, d_n in data:
    dm2 = d_m**2
    dn2 = d_n**2
    diff = dm2 - dn2
    R_i = diff / C
    # instrument uncertainty
    u_inst = math.sqrt((2*d_m*σ_d)**2 + (2*d_n*σ_d)**2)/C
    results.append((m, diff, R_i, u_inst))
results

[(20, 8.679423999999997, 736.4181231970132, 5.0228821184311805),
 (19, 9.269788000000005, 786.5083997963691, 4.8458052429069935),
 (18, 8.530704000000004, 723.799762430002, 4.661472525036131),
 (17, 8.252625000000005, 700.2057525878165, 4.49577451671188),
 (16, 8.203305, 696.0211267605634, 4.334682195521749)]

Python

python
R_values = [r[2] for r in results]
mean_R = statistics.mean(R_values)
std_R = statistics.stdev(R_values)
u_rep = std_R / math.sqrt(len(R_values))
mean_R, std_R, u_rep

(728.5906329543528, 36.41827812711653, 16.286749103145258)

Python

python
u_insts = [r[3] for r in results]
mean_u_inst = statistics.mean(u_insts)
mean_u_inst

4.672123319721587

Python

python
u_c = math.sqrt(u_rep**2 + mean_u_inst**2)
u_c

16.943639888273392

Pensiero

Refining uncertainty calculationRefining uncertainty calculation

I need to compute instrument uncertainties for each pair using the formula: u_R_inst_i = sqrt((2 d_m σ_d)^2 + (2 d_{m-5} σ_d)^2)/C, where σ_d = 0.004 mm. I have d_m values for m = 20 and downwards. After this, I'll average the uncertainties and then combine them. It’s important to determine if I should treat the uncertainties as independent or use a root-sum-square approach when averaging the values. I'll use Python to run these calculations, so we can get precise results.I need to compute instrument uncertainties for each pair using the formula: u_R_inst_i = sqrt((2 d_m σ_d)^2 + (2 d_{m-5} σ_d)^2)/C, where σ_d = 0.004 mm. I have d_m values for m = 20 and downwards. After this, I'll average the uncertainties and then combine them. It’s important to determine if I should treat the uncertainties as independent or use a root-sum-square approach when averaging the values. I'll use Python to run these calculations, so we can get precise results.

Clarifying uncertainty calculation

I’m reconsidering how to calculate the instrument uncertainty when averaging measurements. While I could combine uncertainties using 1/sqrt(N), the user hasn’t explicitly asked for that, instead mentioning the root-sum-square of repeatability and instrument uncertainty. Since they haven’t specified propagating the instrument uncertainty by sqrt(N), I’ll keep it simple and use the average instrument uncertainty (around 4.67 mm) combined with repeatability uncertainty. Let’s go with this approach and proceed with the calculations.

Thought for 48 seconds

Python

python
from docx import Document
from docx.shared import Inches

# Data
data = [
    (20, 5.632, 4.800, 8.6794, 736.4),
    (19, 5.488, 4.566, 9.2698, 786.5),
    (18, 5.277, 4.395, 8.5307, 723.8),
    (17, 5.105, 4.220, 8.2526, 700.2),
    (16, 4.949, 4.036, 8.2033, 696.0)
]

mean_R = 728.6  # mm
u_rep = 16.3    # mm
u_inst = 4.7    # mm
u_c = 16.9      # mm

doc = Document()
doc.add_heading('迈克耳逊圆环法测定曲率半径 R 计算报告', level=1)

doc.add_heading('1. 原始数据', level=2)
table = doc.add_table(rows=1, cols=5)
hdr_cells = table.rows[0].cells
hdr_cells[0].text = 'm'
hdr_cells[1].text = 'd_m (mm)'
hdr_cells[2].text = 'd_{m-5} (mm)'
hdr_cells[3].text = 'd_m^2 - d_{m-5}^2 (mm^2)'
hdr_cells[4].text = 'R_i (mm)'

for row in data:
    row_cells = table.add_row().cells
    for i, item in enumerate(row):
        row_cells[i].text = f'{item}'

doc.add_heading('2. 计算结果', level=2)
doc.add_paragraph(f'平均曲率半径 R̄ = {mean_R:.1f} mm')
doc.add_paragraph(f'重复性不确定度 u_rep = {u_rep:.1f} mm')
doc.add_paragraph(f'仪器不确定度 u_inst = {u_inst:.1f} mm')
doc.add_paragraph(f'合成标准不确定度 u_c = √(u_rep^2 + u_inst^2) = {u_c:.1f} mm')

doc.add_paragraph(f'最终结果： R = ({mean_R:.0f} ± {u_c:.0f}) mm  (k=1)')

file_path = '/mnt/data/曲率半径计算报告.docx'
doc.save(file_path)

file_path

'/mnt/data/曲率半径计算报告.docx'

Risposta

我已按“重复性不确定度 u_rep 与仪器不确定度 u_inst 作平方和再开根号”的方式重新计算，并把全过程整理成 Word 报告：

  • 平均曲率半径 R̄ ≈ 728.6 mm
  • 重复性不确定度 u_rep ≈ 16.3 mm
  • 仪器不确定度 u_inst ≈ 4.7 mm
  • 合成标准不确定度 u_c = √(u_rep^2 + u_inst^2) ≈ 16.9 mm
  • 最终结果：R = (729 ± 17) mm (k = 1)

已生成 .docx 文件，内含原始数据表与计算细节，可直接下载查看：

下载《曲率半径计算报告》

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