Running in the Rain

So, I’ve never heard this myth before, but a friend asked me about it a while ago. The myth states that running in the rain will make you wetter when you arrive at your destination than if you had walked. This was on my mind recently because it was actually a bonus question in the physics lab I’m TAing for this year.

Apparently Mythbusters showed that running is a better option for staying dry, but only after they corrected for a false result they’d obtained in a previous show. So how could you figure this out without going through the hassle (or fun) of running through some rain yourself?

Well, what we could do is set up an idealized situation. Imagine there are 5000 drops of rain falling in every cubic meter above you. Let’s say they fall at 5 m/s straight downwards. To simplify things even further, let’s suppose we ignore the structure of our bodies and just consider a blocky person to have a set width, depth and height. Let’s make up a name for this person; Sponge Bob Square Pants (Bob for short). So, let’s say Bob is 0.25m thick, 0.5m wide, and 1m tall.

If Bob stands in the rain, all of the rain will hit his head. The number of drops hitting him per second is equal to the density of the rain times his width times his height thickness times the velocity of the rain in the downwards direction;

5000 drops/m^3 x 0.25m x 0.5m x 5m/s = 3125 drops/s

If he moves (walks or runs) this amount won’t change because the downward velocity of the rain won’t change with respect to him. But, on the other hand, if Bob walks or runs in the rain, the rain will have a horizontal velocity with respect to him, so the rain will start to hit him in the front. We can find the number of drops hitting his front by the same method.

5000 drops/m^3 x 0.5m x 1m x V = 2500 drops/m x V

I’ve just called Bob’s walking/running speed V. I’ll leave it like that and plug in his speed at the end. If Bob needs to run to his house which is 20m away, a total number of drops of rain will hit him in front and back for the trip which we can calculate. We just need to multiply the above results by the time it takes him to get there and then add them together. That’s just the distance to his house divided by his walking/running speed.

time = 20m/V

But here something funny happens. For the rain hitting his front:

(2500 drops/m x V) x (20m /V) = 2500 drops

Hey! It doesn’t depend on how fast he runs! So it really comes down to the rain hitting his head:

(3125 drops/s) x ( time to get home )…

The faster he runs, the less time it takes him to get home. So to minimize the number of rain drops hitting his head, the faster he needs to run.

… but hey! I’m not the same shape as Sponge Bob, and rain doesn’t always fall straight down, you say?

Well, for those of you who want a more complex analysis, here’s a link to an online running-in-the-rain-wetness calculator. Check it out, it’s fun!

Whatever floats your boat…

Well now. That was a long lasting spontaneous blog hiatus…

Sorry. It’s not that I’ve run out of ideas, it’s mainly lack of time… well, since time is relative, maybe it’s just that I perceive myself as having less time than I actually do. But enough excuses… let’s get back to the physics.

I saw a really neat (3 page!) article on arXiv today called “Google Earth Physics“. I love google, and I love physics, so naturally I had to check it out. The abstract reads,

Google Earth photographs often show ships and their wakes in great detail. We discuss how the images can be used to calculate the velocity of these ships.

Did someone say, do-it-yourself-physics? I knew I had to post this. Naturally, ZapperZ beat me to it… but I’ll go one step further and actually try it out.

The actual article is only three pages long and is very easy to read, so if you want more details I’ll direct you straight to the source. But the main suggestion of the article is that using a very simple formula and making two measurements, you can get a rough estimate of the velocity of a boat. The formula is simply,

boat velocity =  1.25 * Sqrt(wl)/Sin(ang)

; wl is wavelength of wake in meters,

ang is angle between wake and boat direction of motion

The 1.25 comes from some dimensionful constants in front involving Pi and g, but if you use standard units (meters) for the wavelength of the wake, those factors just become 1.25. (The fine print also says this assumes the boat is in deep water. In shallow water this equation isn’t accurate.) You can check out the article for the real formula. To make the measurements you just need to measure an angle and the wavelength shown in the picture up at the top. You can use a protractor for the angle and a ruler for the wavelength and then use the scale given in google maps to convert between your “image length” and the actual length.

I found a boat in my area (toronto) on google maps here. Then I used gimp to measure the wavelength and angle of the wake. I then converted my image length to a real length by measuring the image length of the google scale and a simple ratio:

real distance = wavelength * real length of gscale/img length of gscale

= 76px *  10m / 93px

= 8.17m

And  I measured an angle of around 35 degrees. Plugging that into the equation I get:

boat velocity =  6.23 m/s = 22.4 km/h

I’d say that’s a fair velocity for a boat… but what do I know about boats. Try it for yourself and see if you get the same answer!

Physics Riddle — Of Rope and Wood

I thought it was about time I gave you (yes you!) a physics riddle to go with your morning coffee. I’ll state the problem first and then give a few hints below the fold. Readers with a familiarity with ideas in physics can probably solve this without any hints. If you are unfamiliar with physics, don’t be ashamed to check out the hints. Most importantly: this riddle can be solved without any equations. Feel free to post your solutions in the comments. I’ll give the solution as a comment later on.

Okay. Here it is:

Not to scale...

You have two objects:

• A rope of length a given length.
• Two pieces of (let’s say…) wood joined by a bolt. Together these pieces of wood stretch out to be the same length as the rope.

These objects are the same length, same mass and the same mass per unit length. You now hang them by their endpoints so that they hang side by side (as shown in the picture). The horizontal distance between the endpoints on which they hang is the same for the wood pieces as it is for the rope.

Which object has the lower center of mass?

Note: If you are unfamiliar with the concept of center of mass, check out Rhett’s post on DotPhysics here. You don’t need to understand it all for the purposes of this problem, just know what center of mass is.

Our elevators are more awesome…

I’m a fan of the University of Toronto physics department’s elevators.

…is it because they’re fast?

…do they telepathically know which floor you want before pressing a button?

…do they transport you by moving in more than one dimension?

Nope. Better. Let’s take a look inside…

Hmm… looks pretty standard so fa — but wait! What’s that in the far left corner? It looks like a glass cylinder. Let’s go in for a closer look…

<ghasp>It’s a force meter attached to a 750g weight!</ghasp>

A force meter is a type of measuring instrument that enables you to measure the amount of force acting on the object it is attached to — which in this case is a 750 gram weight.

If you don’t understand why this interesting, you need to understand the following: physicists like solving problems, and at least once in every physicist’s life (s)he wonders “what is the acceleration of this elevator, and what g-force am I experiencing?“. This question, of course, bothers us for the whole 30 second elevator ride, and we wish we had a measuring apparatus to figure it out. Eventually the curiosity subsides and we carry on with our daily lives… but now in the U of T physics department, we don’t have to.

Let me show you how it works. The top of the force meter is attached to the elevator and the bottom is attached to a hanging weight. The force meter will measure the force between the elevator and the weight. If the elevator is not accelerating (even if it is moving at a constant speed) the force between the elevator and the weight will be that of gravity. As the elevator speeds up to move to a higher floor it will have to pull on the weight with a greater force so that it not only counteracts gravity, but also pulls the weight upwards. Newton tells us that the net force acting on an object is its mass times its acceleration. So if we know the force acting on the weight and what its mass is, then we can find out the acceleration. I took a quick reading and noticed that the difference between the force when the elevator was not moving and when the elevator was accelerating upwards was approximately 1 Newton. So we can take this, divide it by 750 grams (using google calculator) and find that the result is an acceleration of: ~ 1.33 m / s².

Great! Now let’s get some context on this. Comparing it to gravity (which is 9.81 m / s²) we can say that an elevator accelerating upwards is equal to a g-force of: 1.14, which is small considering that fighter pilots can withstand a g-force of 9. Taking the elevator downwards will initially give you a g-force of: 0.86, which is roughly comparable to standing on Venus (0.904).

So, yes… very awesome. It makes me wonder how many other universities’ physics departments have little things like this. What kinds of neat publicly accessible physics toys are in your physics department?