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Dad • 11/22/03 • 11:00 PM:

This post is a great setup for one of my favorite short stories from college. One evening in junior year, after catching a late dinner by myself, I was surprised by the rain as I left the South Dining hall. It was coming down pretty hard, and I had no rain gear, so I ran as fast as I could to Walsh Hall (a distance of, perhaps, 200 yards?). When I got to my room, I wondered out loud to my roommate, Pat McRedmond, if the running had done me any good. I conjectured that my forward speed may have caused me to intersect more drops than would have hit me had I walked. Pat was also an engineering student, and without skipping a beat said, “Let’s look at the limiting case. What if you just stood there?” I laughed heartily and considered the problem solved ever since.

Now, I’m forced the rethink the question, at least for a few minutes. First, 2-3 inches per hour is a very heavy rain storm. I can’t imagine it was raining anywhere close to that hard when I conducted my uncontrolled experiment. Second, this problem is ammenable to approximation by first principles, and the process of formulating the problem (which I will not do here) might provide some insight.

For example, we know the terminal velocity and approximate size range of raindrops, so we could estimate the instantaneous number of droplets in the typical human cross-section on the basis of rainfall rate. This is the minimum number of droplets hitting the Flash as he zips through the shower. Walking, of course, offers a smaller vertical cross-section, and reduces (perhaps) the number of drops one intersects due to forward motion. As one’s shoulders get wet, perhaps they become saturated and can no longer hold additional water. It occurs to me that if one were to take quick baby steps one could avoid sticking one’s leg out too far “into the rain.” Also, leaning into the wind might keep you drier because the angle of your body wuld more closely match the angle of the drops. Anyway, the problem could be idealized by considering a 1'x2'x6' rectangular solid being dragged through the rain at varying speeds. No, make that a sponge. We could name this sponge Bob.

Patrick • 11/23/03 • 1:22 AM:

The Flash has a two-doored showed in the middle of his bathroom?

Dan • 11/23/03 • 12:11 PM:

Or, you can consider the insane winds that control South Quad. Take my friend Jake’s experience, when he walked back to Morrissey in the rain and his entire front was soaked, his entire back was dry. There was a line in his grey sweatshirt where the water had and hadn’t hit. it was nearly straight across the top of his shoulders.

Dad • 11/23/03 • 6:24 PM:

I can’t help myself. Must solve this problem! OK, last night as I tried to get to sleep, I couldn’t help trying to formulate this problem. I figured the terminal velocity of a raindrop was about 30 ft/sec. (The hyperphysics site at Georgia State U. gives 29.5 ft/sec, so that was a pretty good guess!) So, for a six-foot-eyed man (oh wait, that’s another problem) it would take about 0.2 seconds for the drop to fall his height. So if he’s (or she’s) walking 5 ft/sec, a drop within 1 ft ahead of him he will hit. The amount of rain hitting him directly from above, will of course depend on his vertical cross-section and the rate of rainfall. Let’s work on that first.

First some assumptions. No wind. Constant (or average) vertical and horizontal cross-sectional areas for the rain walker. Let’s say 2 sq. ft. vertical and 7 sq. ft. horizontal. (Or does the direction refer to the orientation of the blade that slices him? Anyway, you know what I mean.) I think 1 inch per hour is quite enough rain to work with, thank you.

One inch per hour amounts to 0.00144 lb/sec/sq ft. So the amount of rain hitting our subject is a function only of the time he spends in the rain and the area he present to the falling drops. So how much time does it take? For walking, let’s assume 3 mi/hr, which amounts to 1.34 m/sec. This means he will be in the rain for about 75 seconds. So the walker can expect to get 0.216 pounds of water from above.

The runner, of course, will have much less time in the rain, but his running posture might present a greater area to the falling droplets. Let’s assume the runner is not a world-class sprinter and would take about twice as long as an Olympian to sprint the 100 m. So he’s out there about 20 seconds, and if he had the same cross-sectional area, would collect only 0.058 pounds of water. According to these calcs, he would need nearly 4 times the area to get as wet as the walker. But what about the drops that you walk or run into?

I won’t keep you in suspense. The theoretical answer is that it doesn’t matter whether you run or walk, you will intersect the same number of drops! All that matters is that you pass through the same volume of wet air in the course of your walk/run.

At any time, in a steady rain, the amount of water in a given volume of air will be constant. To calculate this, consider that, at 30 ft/sec, it takes 0.033 seconds for each drop to go 1 foot. So a cubic foot of air would contain 0.033x.00144 = 0.000047 pounds of water. The volume swept out by our subject body is 7×328 = 2296 cu. ft. Thus, multiplying these two numbers, our body would hit about 0.11 pounds of water in the course of moving through 100 m of rain. Overall, the walker would receive about a third of a pound of water while the runner would receive only 0.17 pounds, about half as much. If we double the rainfall rate, these numbers would just increase by the same amount. In theory, the runner always stays drier.

Of course, the assumption that the runner and the walker have the same area exposed to the falling rain is not a good one. Clearly, the runner exposes more area as he leans into the running posture and flails his arms and legs. Also, splashing might have a significant unaccounted-for effect. Perhaps most importantly, the walker may cause a great deal of water to run off his shoulders without sticking. I think this process might be exacerbated by wearing rubber suits, as Patrick described.

I don’t know if the original data are available, but I’d be interested in how close my number comes to the actual measured amounts of water. Based on their greater rainfall rate, I would predict between 0.5 and 1 pound of water taken on.

Dad • 11/23/03 • 7:21 PM:

Oh, I forgot, Walk Don’t Run was was a great instrumental hit in the late 50’s by the Ventures.

By the way, Dictionary.com defines “shower” as: A brief fall of precipitation, such as rain, hail, or sleet.

Dad • 11/23/03 • 10:27 PM:

This site has a calculation procedure for determining how wet you will get while walking/running in the rain, under various conditions. It basically confirms my results.

Dad • 11/25/03 • 9:40 AM:

What’s the record for consecutive solo comments? The Discovery channel show was repeated last night, so I decided to tune in. In addition to the rain segment, they “investigated” the exploding toilet myth (wife empties hair spray can into commode; husband throws in lit cigarette and suffers severe trauma when it explodes) and “ice bullet” myth (assasin uses ice bullet to kill target; no trace is found as it melts on impact). They destroyed the other myths, so I think they wanted to destroy the “run in the rain” myth too.

The course was 100 feet long rather than 100 meters. They said the rainfall rate was 2-3 inches per hour and that this was “normal or typical.” Actually, 2-3 in. is a pretty good rainfall in a day. However, the amount of water coming down from the sprinklers looked rather normal to me, so I suspect they got the number wrong. Of course, my calcs say rainfall rate shouldn’t matter in the comparison.

I was surprised at how little water they measured, considering the “torrential” downpour they were moving through. I didn’t get all the numbers, but it appears they were measuring about 30-40 grams weight increase. If I multiply by 3.28 (100 m vs. 100 ft) these numbers fall into the range I calculate for 1 inch/hour rate.

But the real question is why they get wetter running. Well, for one thing, they weren’t running very fast. It looked to me like they were purposely bouncing up and down and moving their arms and legs (to increase the surface area) but not moving forward very rapidly. Everybody knows that when you run in the rain, it’s hell bent for leather and Katie bar the door. I mean, when you turn the corner to get into the building (or whatever) you have to run the risk of slipping on the wet pavement and flopping headlong (or feetlong) into the gutter. Obviously, if you pump your arms and legs, you increase your surface area, so you have to offset that with reduced time in the rain.

Also, these guys weren’t wearing hats or hoods. Did they weigh they water in their hair? Clearly, my idea of using sponge Bob is a better one.

Finally, I’ve thought about the effect of wind, and I think it only increases the amount of rain that gets on you during your stay in the rain. Hence, it increases the advantage to running. Of course if you’re like me and sweat like a racehorse, maybe the perspiration kills any advantage for running.

Dan • 11/25/03 • 8:52 PM:

How about playing Kill the Man with the Ball in rainy, 30 degree nighttime weather? That’s running and walking.

Can we factor in the amount we don’t want to be in the rain in the first place? Isn’t that why you run instead of walk? I ran in the rain the other day and the back of my legs got soaked. I can’t imagine I would have been wetter if I walked.

Dad, SpongeBob isn’t real. He’s fake.

Sometimes I’m proud that you’re my dad. After reading your lab report…

Of course the Flash has a two-doored showed in the middle of his bathroom. It’s right next to the shower, SmartGuy.

David • 10/24/06 • 10:17 PM:

You would think that the New York Times would have something more intelligent and scientific to say on the subject of Run vs. Walk. I guess not. Perhaps they should look at this post.

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