acos(1.000000) = -1.#IND00 ???

[Olumide]

Hi -

I'm computing the angle between pairs of unit vectors as the arc
cosine of the dot product of each pair of vectors. Occasionally, when
the dot product is 1.000000 (output to 6 decimal places) I get the
value -1.#IND00 ! What does this mean?

Thanks,

- Olumide
PS: if it helps, I'm using Visual Studio .NET 2003 on a Windows XP
machine.

[=?UTF-8?B?RXJpayBXaWtzdHLDtm0=?=]

On 2007-10-14 21:33, Olumide wrote:

Hi -

I'm computing the angle between pairs of unit vectors as the arc
cosine of the dot product of each pair of vectors. Occasionally, when
the dot product is 1.000000 (output to 6 decimal places) I get the
value -1.#IND00 ! What does this mean?

It probably means that one of the decimals that you have not shown is
greater than 1. acos is only defined for values in the range -1 to 1
inclusive, for other values of input a domain error occurs and it is up
to the implementation what will be returned. In VC++ this produces a so
called quiet NaN (Not A Number) which looks like -1.#IND when printed.

--
Erik Wikström

[=?UTF-8?B?RXJpayBXaWtzdHLDtm0=?=]

On 2007-10-14 22:13, Erik Wikström wrote:

On 2007-10-14 21:33, Olumide wrote:

>Hi -

I'm computing the angle between pairs of unit vectors as the arc
cosine of the dot product of each pair of vectors. Occasionally, when
the dot product is 1.000000 (output to 6 decimal places) I get the
value -1.#IND00 ! What does this mean?

It probably means that one of the decimals that you have not shown is
greater than 1.

Sorry, that should be "greater than 0".

--
Erik Wikström

[Olumide]

On 14 Oct, 21:24, Erik Wikström <[e-mail]:

On 2007-10-14 22:13, Erik Wikström wrote:

I'm computing the angle between pairs of unit vectors as the arc
cosine of the dot product of each pair of vectors. Occasionally, when
the dot product is 1.000000 (output to 6 decimal places) I get the
value -1.#IND00 ! What does this mean?

It probably means that one of the decimals that you have not shown is
greater than 1.

Sorry, that should be "greater than 0".

Thanks, I should have known that ;-) .