# [time-nuts] How can this be?

Poul-Henning Kamp phk at phk.freebsd.dk
Tue Oct 19 06:35:12 EDT 2004

```In message <4174EA5B.2090407 at usa.net>, Alberto di Bene writes:

97.22 ns - 69.44 ns = 27.78 nsec :-)

Remember that he measures the phase of one against the other
and finds the phase range to be 250-350 degrees.  Some constant
part of this is made up of cable-lengths and other (mostly) constant
factors, what's interesting is the 100 degrees span which works
out to roughly 27nsec.

Poul-Henning

>
>   Err, please correct me if I am wrong, but given that the period of 10
>   MHz is
>   100 ns, wouldn't 250 degrees correspond to 100 * 250/360 = 69.44 ns
>   and similarly 350 degrees => 100* 350/360 = 97.22 ns ?
>   Still inside specs.
>   73  Alberto  I2PHD
>   ---------------------------------------------
>   Poul-Henning Kamp wrote:
>
>In message [1]<009501c4b56f\$256e4f00\$0500a8c0 at darius.domain.actdsltmp>, "Bill H
>awk
>ins" writes:
>
>
>Have two Z3801A receivers by HP and a Racal-Dana model 1992
>counter timer with a Phase A-B function. Measured the phase
>between the two 10 MHz outputs of the receivers and got
>values between 250 and 350 degrees with 2-3 degrees noise.
>
>
>That's 27 nsec according to my math and that's inside spec.
>
>References
>
>   1. mailto:009501c4b56f\$256e4f00\$0500a8c0 at darius.domain.actdsltmp
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>

--
Poul-Henning Kamp       | UNIX since Zilog Zeus 3.20
phk at FreeBSD.ORG         | TCP/IP since RFC 956
FreeBSD committer       | BSD since 4.3-tahoe
Never attribute to malice what can adequately be explained by incompetence.

```