Experimental Work №3

Subject. Measuring the size of the small objects

Theoretical data and practical advice

Sometimes the device’s scale value does not allow measurement to be made with sufficient accuracy. For example, when the device’s scale value is larger or comparable to the size to be measured. In this case, apply the string method. The row method allows us to determine only the mean value of a small object.

To determine the size d of a small object b with the string method, it is necessary:

  1. Form the row, for example, lay the seeds in a row, close to each other;
  2. Measure the length \(L\) of the row;
  3. Determine the number \(n\) of objects or turns in a row;
  4. 4. Find the ratio: \(d = \frac{L}{n}\).

Figure 1

Subject. Measuring the size of the small objects.

Target: to determine the diameter of a pea, the diameter of millet grain, the thickness of the thread by the string method.

Equipment: ruler (treadmill), threads, a metal rod, glasses with peas and millet.


The results of the measurements I’m writing down into the table:

Table 1

No. of the experiment Object Length \(L\) of the row, mm Amount \(n\) of objects (turns) in a row Diameter (thickness) \(d_{meas}, mm\) Absolute error \(\Delta d, mm\) Relative error
\(\varepsilon,\) %

1. I’m determining the scale division of the ruler (treadmill): $$ C_{rul} = \frac{\FormInput{name}}{\FormInput{name}} = \ \FormInput{name} \ mm $$

Figure 2

2. I’m making a row of peas, measure the length \(L_1\) of the row, amount \(n_1\) of peas in a row and calculating \(d_1\) the peas diameter: $$d_1 = \frac{L_1}{n_1}$$ $$d_1 = \frac{\FormInput{name}}{\FormInput{name}} = \FormInput{name} \ mm$$

Figure 3

3. I’m making a row of millet grains, measure the length \(L_2\) of the row, amount \(n_2\) of grains in a row and calculate \(d_2\) the diameter of a millet grains: $$d_2 = \frac{L_2}{n_2}$$ $$d_2 = \frac{\FormInput{name}}{\FormInput{name}} = \FormInput{name} \ mm$$

Figure 4

4. In order to form a row, I wind the thread on the metal rod, measure the length \(L_3\) of the row, number \(n_3\) of turns and calculate the average thread thickness \(d_3\) $$d_3 = \frac{L_3}{n_3}$$ $$d_3 = \frac{\FormInput{name}}{\FormInput{name}} = \FormInput{name} \ mm$$

5. I’m estimating the absolute error. in this case the absolute error is less, than the scale division by as many times as many objects (turns) in a row: $$ \Delta d_1 = \frac{1 mm}{n_1}; \, \Delta d_1 = \frac{\FormInput{name}}{\FormInput{name}} = \FormInput{name} \ mm; d_1 = \FormInput{name} \pm \FormInput{name}; $$ $$ \Delta d_2 = \frac{1 mm}{n_2}; \, \Delta d_2 = \frac{\FormInput{name}}{\FormInput{name}} = \FormInput{name} \ mm; d_2 = \FormInput{name} \pm \FormInput{name}; $$ $$ \Delta d_3 = \frac{1 mm}{n_3}; \, \Delta d_3 = \frac{\FormInput{name}}{\FormInput{name}} = \FormInput{name} \ mm; d_3 = \FormInput{name} \pm \FormInput{name}; $$

6. I make an estimate of the measurement results of relative error: $$ \varepsilon_1 = \frac{\Delta d_1}{d_1}; \, \varepsilon_1 = \frac{\FormInput{name}}{\FormInput{name}} = \FormInput{name} \%; $$ $$ \varepsilon_2 = \frac{\Delta d_2}{d_2}; \, \varepsilon_2 = \frac{\FormInput{name}}{\FormInput{name}} = \FormInput{name} \%; $$ $$ \varepsilon_3 = \frac{\Delta d_3}{d_3}; \, \varepsilon_3 = \frac{\FormInput{name}}{\FormInput{name}} = \FormInput{name} \%; $$

7. The results of the experiment:


Experiment #3 (Mechanics)
Experimentator: test (guest@guest.uaa)

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