1. Clarke 1880 ellipsoid has a semi-major axis of \( a = 6378249.145 \) m and inverse flattening \( 1/f = 293.465 \). Compute all the ellipsoidal parameters \( b \), \( e \), and \( e' \).

Given:
\( a = 6378249.145 \) m
\( 1/f = 293.465 \)

We can calculate \( b \) using the formula:
\[ b = a(1 - 1/f) \]

Substitute the given values:
\[ b = 6378249.145(1 - 1/293.465) \]
\[ b = 6378249.145 \times 0.996594 \]
\[ b = 6356863.016 \text{ m} \]

Next, we can calculate the first eccentricity \( e \) using the formula:
\[ e = \sqrt{\frac{(a^2 - b^2)}{a^2}} \]

Substitute the values:
\[ e = \sqrt{\frac{(6378249.145^2 - 6356863.016^2)}{6378249.145^2}} \]
\[ e = \sqrt{\frac{(40680631590750.102 - 40320736203569.898)}{40680631590750.102}} \]
\[ e = \sqrt{\frac{359195187180.204}{40680631590750.102}} \]
\[ e = \sqrt{0.008831} \]
\[ e = 0.094 \]

Finally, we can calculate the second eccentricity \( e' \) using the formula:
\[ e' = \sqrt{\frac{(a^2 - b^2)}{b^2}} \]

Substitute the values:
\[ e' = \sqrt{\frac{(6378249.145^2 - 6356863.016^2)}{6356863.016^2}} \]
\[ e' = \sqrt{\frac{359195187180.204}{40320736203569.898}} \]
\[ e' = \sqrt{0.008907} \]
\[ e' = 0.094 \]

Therefore, the ellipsoidal parameters are:
\( b = 6356863.016 \) m
\( e = 0.094 \)
\( e' = 0.094 \)

SOLUTION

Ellipsoidal parameters can be computed as follows:

$$ b = a\left( 1 - \frac{1}{f} \right); \quad e^2 = \frac{a^2 - b^2}{a^2}; \quad e'^2 = \frac{a^2 - b^2}{b^2} $$

Plugging the numbers, we get:

$$ b = 6356514.966 \ \text{m} $$

$$ e^2 = \frac{6378249.145^2 - 6356514.966^2}{6378249.145^2} = 0.0066943 $$

$$ e'^2 = \frac{6378249.145^2 - 6356514.966^2}{6356514.966^2} = 0.0067385 $$

Therefore:

$$ e = 0.08181919 $$

$$ e' = 0.08227195 $$