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 ρ = 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 = ρ*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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From: Phys-l <phys-l-bounces@mail.phys-l.org> on behalf of David Bowman via
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Subject: Re: [Phys-L] raining
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Assuming a temporally and spatially constant/uniform density of rain water in
the air (and an upright posture) then the amount of rain striking your head and
shoulders (and any surface with a horizontal projection) from above will be
proportional to the time you spend out in the rain. But the amount of rain
striking your front side (and any surface with a front facing vertical
projection) will be constant independent of time (within reason) since that
amount of water is just the constant amount of rain in the volume of space you
must sweep through on your way through the rain laden air to get to where you
are going. Thus, your front side gets just as wet no matter how fast you move
through the rain. But your head and shoulders will get the least wet if the
pass through the rainy region as fast as possible.
David Bowman
-------- Original message --------
From: Anthony Lapinski via Phys-l <phys-l@mail.phys-l.org>
Date: 12/19/23 10:33 PM (GMT-05:00)
To: John Denker via Phys-l <phys-l@mail.phys-l.org>
Cc: Anthony Lapinski <alapinski@pds.org>
Subject: [Phys-L] raining
CAUTION: This email originated from outside Georgetown College's Email System.
DO NOT CLICK on any links or open attachments unless you recognize the sender
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Trying to get a "scientific" answer to the question: will you get more wet
if you walk or run in the rain? Conflicting results/videos online, and some
consider wind and leaning bodies. I just want the basic parameter that the
rain is falling straight down and the person is moving "upright."
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