Re: [Phys-L] raining
Seems to me, always has, that there is nothing (that HAS to get) complicated
about this question. Just consider the limits. Move at zero speed and you
get infinitely wet from the top down. Let the Flash run through the rain
and his head is essentially dry, but he did have to run through the volume
of rain in the 3 dimensional path cut out of the essentially stationary rain
in his way. But that amount in finite and as David has pointed out, fast or
slow you will intercept that amount of water to your front. Running will
always collect the least amount of moisture on the person.
Richard W. Tarara
Professor of Physics, emeritus
Saint Mary?s College
Notre Dame, Indiana
Free Physics Instructional Software
Windows and Mac
sites.saintmarys.edu/~rtarara/software.html
-----Original Message-----
From: Phys-l <phys-l-bounces@mail.phys-l.org> On Behalf Of David Bowman via
Phys-l
Sent: Wednesday, December 20, 2023 8:29 AM
To: Phys-L@Phys-L.org
Cc: David Bowman <David_Bowman@georgetowncollege.edu>
Subject: Re: [Phys-L] raining
Weird. Each Greek [rho] in my previous post got translated into a tilded n
after passing through the Phys-l list. I trust the readers can cope with
that. It's a good thing I didn't use more than one Greek letter.
David Bowman
________________________________________
From: Phys-l <phys-l-bounces@mail.phys-l.org> on behalf of David Bowman via
Phys-l <phys-l@mail.phys-l.org>
Sent: Wednesday, December 20, 2023 8:16 AM
To: Phys-L@Phys-L.org
Cc: David Bowman
Subject: Re: [Phys-L] raining
Elaborating further on my previous post.
Assume the walking/running direction is horizontal/perpendicular to the
vertically falling rain & assuming the falling rain has a uniform
distribution in space and time.
Let ñ = mass density of falling rain (per unit volume of air + rain).
Let L = horizontal distance walked/ran through in the rain.
Let v_t = (mass weighted) mean terminal velocity of falling raindrops Let v
= speed of walking/running Let A_th = body's projected top facing horizontal
cross section area.
Let A_fv = body's projected front facing vertical cross section area.
Let W = mass of water intercepted by moving body.
Then W = ñ*L*(A_fv + (v_t/v)*A_th) .
This is just the projected gross rain interception rate due to simple
kinematics. It obviously neglects the hydrodynamics of tiny deflected air
motions partially carrying drops in the immediate vicinity of body
collisions with drops causing altered localized trajectories of drops that
could affect the collision rate by a tiny amount.
But a much bigger neglected effect is due to wind. Wind causes a number of
complications. Here are 4 of them off the top of my head. First, wind
tends to be gusty & quite time dependent and thus hard to model. Second,
wind changes the direction of the falling rain so the direction of the
projected area perpendicular to it must also change. Third, because of the
poly-disperse distribution of drop sizes there is a distribution of
different terminal velocities of the drops and this causes different sized
drops to fall/move in different directions at different speeds. Fourth,
there is a Galilean transformation involved for the effective walking speed
due to any component of wind which may be parallel the walking direction,
and this changes the effective walking speed for those drops which are
carried with that horizontal component of velocity by the wind.
I suspect there are even more complications from wind, but I'm not in the
mood to try to think of them.
David Bowman
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