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Is not "phrased in an elaborate way"; it is expressed in the current (at that point in time) theory of ratios, when the symbolic algebra was still in his infancy.

Uniform speed is defined in Two new sciences, Third Day :

By steady or uniform motion, I mean one in which the distances traversed by the moving particle during any equal intervals of time, are themselves equal.

This is exactly the current definition of uniform motion : if $t_1=t_2$ then $s_1=s_2$, that implies : $v= \dfrac s t = \text {const}$.

Some "obvious" axioms are set forth and then some theorems are proved :

Th.I If a moving particle, carried uniformly at a constant speed, traverses two distances the time-intervals required are to each other in the ratio of these distances.

Th.III In the case of unequal speeds, the time-intervals required to traverse a given space are to each other inversely as the speeds.

I.e., given space $s$ :

$\dfrac {t_1} {t_2} = \dfrac {v_1} {v_1}$

which amount to :

$v_1 \times t_1 = v_2 \times t_2$.

And also :

Th.IV If two particles are carried with uniform motion, but each with a different speed, the distances covered by them during unequal intervals of time bear to each other the compound ratio of the speeds and time intervals.

I.e.:

$\dfrac {v_1} {v_2} = \dfrac {s_1} {s_2} \times \dfrac {t_2} {t_1}$.

As you can easily verify, it is :

$\dfrac {v_1} {v_2} = \dfrac {s_1} {t_1} \times \dfrac {t_2} {s_2}$.

Is not "phrased in an elaborate way"; it is expressed in the current (at that point in time) theory of ratios, when the symbolic algebra was still in his infancy.

Uniform speed is defined in Two new sciences, Third Day :

By steady or uniform motion, I mean one in which the distances traversed by the moving particle during any equal intervals of time, are themselves equal.

This is exactly the current definition of uniform motion (i.e. constant speed) : if $t_1=t_2$ then $s_1=s_2$, that implies : $v= \dfrac s t = \text {const}$ and yes, it is an "idealization" as anything in science.

Some "obvious" axioms are set forth and then some theorems are proved :

Th.I If a moving particle, carried uniformly at a constant speed, traverses two distances the time-intervals required are to each other in the ratio of these distances.

I.e. with $v= \text {const}$ :

$\dfrac {t_1} {t_2} = \dfrac {s_1} {s_2}$, that amounts to : $\dfrac {s_1} {t_1} = \dfrac {s_2} {t_2}$.

And :

Th.III In the case of unequal speeds, the time-intervals required to traverse a given space are to each other inversely as the speeds.

I.e., given space $s$ :

$\dfrac {t_1} {t_2} = \dfrac {v_1} {v_1}$, which amount to : $v_1 \times t_1 = v_2 \times t_2$.

And also :

Th.IV If two particles are carried with uniform motion, but each with a different speed, the distances covered by them during unequal intervals of time bear to each other the compound ratio of the speeds and time intervals.

I.e.:

$\dfrac {v_1} {v_2} = \dfrac {s_1} {s_2} \times \dfrac {t_2} {t_1}$.

As you can easily verify, it is :

$\dfrac {v_1} {v_2} = \dfrac {s_1} {t_1} \times \dfrac {t_2} {s_2}$.

Is not "phrased in an elaborate way"; it is expressed in the current (at that point in time) theory of ratios, when the symbolic algebra was still in his infancy.

Uniform speed is defined in Two new sciences, Third Day :

By steady or uniform motion, I mean one in which the distances traversed by the moving particle during any equal intervals of time, are themselves equal.

Some "obvious" axioms are set forth and then some theorems are proved :

Th.I If a moving particle, carried uniformly at a constant speed, traverses two distances the time-intervals required are to each other in the ratio of these distances.

Th.III In the case of unequal speeds, the time-intervals required to traverse a given space are to each other inversely as the speeds.

I.e., given space $s$ :

$\dfrac {t_1} {t_2} = \dfrac {v_1} {v_1}$

which amount to :

$v_1 \times t_1 = v_2 \times t_2$.

And also :

Th.IV If two particles are carried with uniform motion, but each with a different speed, the distances covered by them during unequal intervals of time bear to each other the compound ratio of the speeds and time intervals.

I.e.:

$\dfrac {v_1} {v_2} = \dfrac {s_1} {s_2} \times \dfrac {t_2} {t_1}$.

As you can easily verify, it is :

$\dfrac {v_1} {v_2} = \dfrac {s_1} {t_1} \times \dfrac {t_2} {s_2}$.

Is not "phrased in an elaborate way"; it is expressed in the current (at that point in time) theory of ratios, when the symbolic algebra was still in his infancy.

Uniform speed is defined in Two new sciences, Third Day :

By steady or uniform motion, I mean one in which the distances traversed by the moving particle during any equal intervals of time, are themselves equal.

This is exactly the current definition of uniform motion : if $t_1=t_2$ then $s_1=s_2$, that implies : $v= \dfrac s t = \text {const}$.

Some "obvious" axioms are set forth and then some theorems are proved :

Th.I If a moving particle, carried uniformly at a constant speed, traverses two distances the time-intervals required are to each other in the ratio of these distances.

Th.III In the case of unequal speeds, the time-intervals required to traverse a given space are to each other inversely as the speeds.

I.e., given space $s$ :

$\dfrac {t_1} {t_2} = \dfrac {v_1} {v_1}$

which amount to :

$v_1 \times t_1 = v_2 \times t_2$.

And also :

Th.IV If two particles are carried with uniform motion, but each with a different speed, the distances covered by them during unequal intervals of time bear to each other the compound ratio of the speeds and time intervals.

I.e.:

$\dfrac {v_1} {v_2} = \dfrac {s_1} {s_2} \times \dfrac {t_2} {t_1}$.

As you can easily verify, it is :

$\dfrac {v_1} {v_2} = \dfrac {s_1} {t_1} \times \dfrac {t_2} {s_2}$.

Is not "phrased in an elaborate way"; it is expressed in the current (at that point in time) theory of ratios, when the symbolic algebra was still in his infancy.

For uniform speed :

$\dfrac {v_1} {v_2} = \dfrac {s_1} {s_2} \times \dfrac {t_2} {t_1}$.

As you can easily verify, it is :

$\dfrac {v_1} {v_2} = \dfrac {s_1} {t_1} \times \dfrac {t_2} {s_2}$.

For unequal speeds :

given the space $s, \ \dfrac {t_1} {t_2} = \dfrac {v_1} {v_1}$

which amoiunt at :

$v_1 \times t_1 = v_2 \times t_2$.

Is not "phrased in an elaborate way"; it is expressed in the current (at that point in time) theory of ratios, when the symbolic algebra was still in his infancy.

Uniform speed is defined in Two new sciences, Third Day :

By steady or uniform motion, I mean one in which the distances traversed by the moving particle during any equal intervals of time, are themselves equal.

Some "obvious" axioms are set forth and then some theorems are proved :

Th.I If a moving particle, carried uniformly at a constant speed, traverses two distances the time-intervals required are to each other in the ratio of these distances.

Th.III In the case of unequal speeds, the time-intervals required to traverse a given space are to each other inversely as the speeds.

I.e., given space $s$ :

$\dfrac {t_1} {t_2} = \dfrac {v_1} {v_1}$

which amount to :

$v_1 \times t_1 = v_2 \times t_2$.

And also :

Th.IV If two particles are carried with uniform motion, but each with a different speed, the distances covered by them during unequal intervals of time bear to each other the compound ratio of the speeds and time intervals.

I.e.:

$\dfrac {v_1} {v_2} = \dfrac {s_1} {s_2} \times \dfrac {t_2} {t_1}$.

As you can easily verify, it is :

$\dfrac {v_1} {v_2} = \dfrac {s_1} {t_1} \times \dfrac {t_2} {s_2}$.