I get the correct result. From the help file I understand that if any of the arguments is a floating number the result should be a floating number. But that does not seem the case above so apparently I am missing something here. Can anybody explain the logic behind the above?

I get the correct result. From the help file I understand that if any of
the arguments is a floating number the result should be a floating number.
But that does not seem the case above so apparently I am missing something
here. Can anybody explain the logic behind the above?

In the first case, 6371007^2 is calculated using integer arithmetic,
and the result (40589730194049) will exceed the range of an "int"
(2^31-1 = 2147483647), typically resulting in overflow (R^2 will
probably equal -2005720447). The value will be converted to "double"
for the multiplication, but by that point the error has already
occurred.

In the second case, R is converted to "float" (single-precision
floating-point), and the result of squaring it is well within the
range of a "float", although it will be rounded to 40589731037184.0
(using double() instead of float() will use double-precision and avoid
the rounding error).

Thanks for the explanation. Good to know that the problem lies with the exceeding the integer range… something I have to be careful about in the future.

I get the correct result. From the help file I understand that if any of
the arguments is a floating number the result should be a floating number.
But that does not seem the case above so apparently I am missing something
here. Can anybody explain the logic behind the above?

In the first case, 6371007^2 is calculated using integer arithmetic,
and the result (40589730194049) will exceed the range of an “int”
(2^31-1 = 2147483647), typically resulting in overflow (R^2 will
probably equal -2005720447). The value will be converted to “double”
for the multiplication, but by that point the error has already
occurred.

In the second case, R is converted to “float” (single-precision
floating-point), and the result of squaring it is well within the
range of a “float”, although it will be rounded to 40589731037184.0
(using double() instead of float() will use double-precision and avoid
the rounding error).