Heavy vs light person running into the wind

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Carson Dyle

Diamond Member
Jul 2, 2012
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Lighter person, all other things being equal, same height (or even the same person,) 25 lbs lighter, is going to be moving less mass _and_ he's going to have the lower coefficient of drag. So it's almost a moot question. The lighter person has it easier on both fronts.

But I still contend that being 17% lighter is going to make the task easier, far more than whatever difference in drag the two bodies have. Take a more extreme example: 5'8" 150 lb runner vs. 6'4" 175 lb runner. Lighter, average body type vs heavier, thin body. I'll contend again that the lighter guy has an easier time, whatever the wind resistance. I doubt that there's even much difference in the coefficient of drag between the two, and not enough to make much difference.
 

mrjminer

Platinum Member
Dec 2, 2005
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He did say they were roughly the same size and physical ability (proportional to weight, the same, is the impression I got). So, the only factor we should take into account is the weight? I am going with the heavier person.

Everyone else seems to go with the lighter person. No other reason in particular.
 

KIAman

Diamond Member
Mar 7, 2001
3,342
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First, you have to define what you mean by "difficulty."

Does it mean the calories burned? Does it mean the subjective feeling of how hard it is?

Assuming couple of things.

1. Running on the surface of the earth
2. Both have the same type of running shoes with the same friction
3. Both run the same speed
4. Both have the same gait when running

With these assumptions, the heavier person will expend more calories not because of the horizontal distance and speed they ran against the wind but because of the slight vertical motion from the gait of the run working against gravity. Even if both ran without any variance in the vertical plane, the required movement of the mass of the (heavier) legs and arms will cause additional expended calories.

Edit: In the end, common sense could have solved this with a quick guess.
 
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SKORPI0

Lifer
Jan 18, 2000
18,470
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Why would a Wookiee, an 8-foot-tall Wookiee, want to live on Endor, with a bunch of 2-foot-tall Ewoks? That does not make sense! But more important, you have to ask yourself: What does this have to do with this case? ............, it does not make sense! If Chewbacca lives on Endor, you must acquit! The defense rests.

tumblr_m3dqhkdd8e1r7mvdfo1_500.jpg

:confused::confused::confused:....................:biggrin::biggrin::biggrin:
 

pete6032

Diamond Member
Dec 3, 2010
7,972
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He did say they were roughly the same size and physical ability (proportional to weight, the same, is the impression I got). So, the only factor we should take into account is the weight? I am going with the heavier person.

Everyone else seems to go with the lighter person. No other reason in particular.
Theoretically, say they were the exact same size person, one just happens to weight 25 pounds more. Running into the wind would be no more difficult for the heavier person over the lighter person, if we discount the normal additional exertion associated with carrying the extra 25 lbs under zero wind.?
 

vi edit

Elite Member
Super Moderator
Oct 28, 1999
62,484
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Theoretically, say they were the exact same size person, one just happens to weight 25 pounds more. Running into the wind would be no more difficult for the heavier person over the lighter person, if we discount the normal additional exertion associated with carrying the extra 25 lbs under zero wind.?

Now things get really complicated and it really just comes down to the actual ability to output force. If shapes are identical, then in theory the person that is heavier should be less impacted. Wind just wont manipulate their forward momentum as much. But that's assuming they are capable of properly carrying that weight in the first place.
 

tynopik

Diamond Member
Aug 10, 2004
5,245
500
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if they're the same size and only difference is density:

ACTUAL extra energy expended by each person is the same as the force imparted by the wind is the same (based solely on coefficient of drag)

RELATIVE extra energy will be less for the heavy person because they're expending more energy so a fixed amount is relatively smaller for them

think of it like hauling a trailer behind a VW bug vs hauling the same trailer behind a F-350

The actual energy required to tow the trailer is the same, but the impact on the bug is more.
 

DigDog

Lifer
Jun 3, 2011
14,368
2,831
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No, he doesn't. The human body is mostly water and water weight is fixed. Unless one person has a lead skeleton the heavier person is going to have greater surface area and a greater coefficient of drag. If he wanted to use identical shapes of different weights he would not have specified people.

reality does not matter to math. also, we do have such think as elements of different density, so with a rewording you could have the same example and be correct too.

or, you would rather be the one guy who

"if i give you one cat, and then another cat, how many cats do you have?"

"three - i already have a cat"
 

DigDog

Lifer
Jun 3, 2011
14,368
2,831
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jesus christ of the holy physics, i'm seeing some dumb **** here

1) TWO OBJECTS HAVE IDENTICAL SHAPE BUT DIFFERENT DENSITY

2) THEY BOTH HAVE AN IDENTICAL VECTOR

3) how would (A CONSTANT) wind affect them
 
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