kinetic energy – Thoughts on Energy

Melting Snow while Backpacking

Introduction

I enjoy backpacking in the great outdoors. As all good backpackers know, planning for access to water is a critical part of any backpacking trip. Backpacking is a lot of work and usually involves a lot of sweating, so keeping hydrated is necessary. However, maintaining a backpack as light as possible is also a key consideration, so one doesn’t want to carry a lot of water around. Thus, the usual strategy is to stay close to natural sources of water and bring a good water filter to allow for regular refills of water containers.

As a side note, how much work is a backpacking trip, you ask? Oh, at 100% efficiency, almost one million Joules worth of work would be required on a typical mountainous trail with about 1,000 meters of up and down (average 5% grade for a 20 kilometer, or about 12.5 mile, trip) during the full trip. (That calculation is simply the potential energy, which is the mass multiplied by acceleration due to gravity multiplied by the total elevation change). However, considering the body is probably only about 5% efficient in this case, the real energy use is approximately 20 million Joules, which is equivalent to a total of 5,000 food calories. That is just the extra calories of food energy “burned” during the trip in addition to the resting metabolic calorie burn of about 2,000 calories per day.

While on a backpacking trip some time ago, my brother and I took a day hike up a mountain and we knew we would not be near any sources of water for a good portion of the day. I had two water bottles with me on the backpacking trip, one un-insulated bottle and one insulated Thermos bottle. However, I didn’t have enough foresight to bring a smaller bag on the trip to carry around for the day hike, so all I had was a plastic grocery sack (I know, some Californians reading this post will cringe at those words), and thus I only took one of my water bottles. Unfortunately, it turned out that I selected the wrong water bottle to take with me.

You see, even though I tried to ration my water intake, I found I was out of water with several miles left to hike. Fortunately, we found some snow that I ate a good amount of, and also packed a good amount into my water bottle. I figured it could melt as we went along our merry way and I would soon have some nice cool water to drink. Well, that didn’t work out as well as I hoped because the snow was painfully slow to melt. Ironically, as we continued our hike, we ended up in a downpour that lasted a few hours. So I was soaked to the bone and dehydrated at the same time, and I still couldn’t get the darn snow to melt in my water bottle!

I had plenty of time during the hike to think about methods to melt the snow into water so I could drink it. Since it relates to energy, I thought I would share my findings here. The underlying assumption below is that conduction heat transfer through the insulated water bottle was negligibly low in the short term, which I think turned out to be a pretty good assumption. Also, a spreadsheet containing the calculations is available below.

Air Blowing Method

My first thought was to simply blow air into the water bottle’s straw and have it come out the little pressure release hole. It would work like any basic heat exchanger. I could transfer heat from the air I exhaled into the snow and cause it to melt. We should all be somewhat familiar with heat exchangers, as many of us have at least two in/near our dwelling places as part of an air conditioning system. One heat exchanger (located inside) allows the transfer of heat from the air inside the living space into a refrigerant, and the other heat exchanger (located outside) allows the transfer of heat from the refrigerant to the outside air.

We can use a simple equation to determine how much air I would need to blow through my water bottle straw to melt the snow. To formulate this equation, I simply use the fact that the amount of heat required to melt the snow must be the same as the amount of heat lost by the air as it travels through the snow in the bottle.

Where

Q is the total amount of heat required to be transferred from the air to the snow, in Joules.
is the total mass of air required, in kilograms.
is the specific heat of air, which is 1.0 Joules per kilogram degrees Celsius.
is the change in temperature experienced by the air as it moves from my mouth to the exit hole, in degrees Celsius. I assume the air temperature goes from about 32 degrees Celsius when it exits my mouth to about 5 degrees Celsius when it exits the water bottle, for a change in temperature of 27 degrees C.
is the total mass of snow, in kilograms. My water bottle fits about 0.5 liters of water. I haven’t packed the snow in completely tight, so I have about 0.4 liters of snow at a density of about 0.6 kilograms per liter, for a total of 0.24 kilograms of snow.
is the latent heat of fusion ((Wikipedia link)) of water, which is 334,000 Joules per kilogram.

Now all we need to do is solve for to obtain how much air it would take going through the “heat exchanger” to melt the snow. However, just knowing the total mass of air required is not very helpful. What I really want to know is how long it will take to get that mass of air through the water bottle. Thus, I need to know my respiration rate. I need to know how much air (in units of mass, or kg) I can blow through the bottle in a given amount of time. A typical adult respiration rate is about 6 liters of air per minute. However, when I am hiking with a backpack on I am exercising relatively vigorously, so I probably have a respiration rate around 50 liters per minute.

At an altitude of about 6,000 feet above sea level, the density of air is about 1.0 kilograms per liter. Using all of this information and solving for the time required to melt the snow (see the spreadsheet attached below if you want to follow all the math there), I obtain:

So it would take me about an hour to get enough air through the water bottle to melt the snow. However, this assumes that all of the air I exhale goes through the water bottle. I found that to be unsustainable in practice, as the water bottle restricted my exhalations enough that I would quickly run out of breath (remember, I am hiking on mountains wearing a 35-pound backpack). So I could really only get about one third of the air I exhaled through the water bottle. That results in about three hours to melt the snow. I am not that patient.

Potential Energy Input Method

My next thought was to use potential energy of the snow to melt it. The thought process is relatively simple. When I held the water bottle upright, there was a small pocket of air at the top. So if I turned the water bottle over, there would be stored potential energy in the lump of snow because the snow tended to stick to the bottom of the bottle.

As you will recall, potential energy is stored mechanical energy. In this case, it is the energy of the snow that can be realized as the snow falls the short distance from one end of the water bottle to the other. That potential energy must go somewhere as the snow falls. Initially, some energy becomes kinetic energy as the snow moves. When the snow stops against the bottom end of the bottle, about the only place for that energy to go is in the form of heat to the snow. Bingo! All I have to do is tip the water bottle upside-down enough times and I convert enough potential energy into heat energy to melt the snow!

So how many times do I need to invert the water bottle? Using the equation from the previous example, I already know it takes about 80,000 Joules of energy to melt the snow. So I just need to figure out the amount of potential energy for the snow in one time of inverting the water bottle.

It turns out the potential energy is about 0.1 Joules. So I only need to invert the water bottle 850,000 times. Assuming I can invert the water bottle once per second, this ends up being about 10 days. I am definitely not that patient.

Shaking Method

So what if I could speed up the potential energy process with some kinetic energy instead? Instead of inverting the bottle and waiting for the snow to fall in the water bottle, I can just orient the bottle horizontally and shake it back and forth instead and get a similar effect. I can increase the rate of impacts between the snow and the inside of the water bottle and maybe even increase the amount of heat added to the snow on each impact.

Consider the physics of what is going on inside the bottle as I shake it back and forth. As I shake the bottle in one direction, the bottle and snow move together. Then as I move the bottle in the other direction, the snow slams into one end of the bottle. This motion can be represented as a sinusoidal function. All I need to know are the amplitude of the motion and the period of the motion and I can fully describe the maximum velocity and acceleration of the motion. I assume the total range of motion is about 20 centimeters and I can move the bottle back and forth about three times per second. Thus, the motion of the bottle can be represented in this plot:

The position, velocity, and acceleration of the bottle can be represented by these equations (with just some calculus magic going on in the background):

In these equations, A is the amplitude of the motion, which is 10 centimeters, and T is the period of the motion, or the time it takes to complete a full cycle of motion, which is 0.333 seconds. The maximum velocity of the bottle (represented by the first portion of the velocity equation) is about 2 meters per second and the maximum acceleration (again, represented by the first portion of the acceleration equation) is about 35 meters per second squared, or about 3.5 times the acceleration of gravity. Considering a major league baseball pitcher can move a baseball from zero to 95 miles per hour in about 0.3 seconds (acceleration of about 15 times that of gravity), that seems like a reasonable value for acceleration.

It is the maximum velocity that I am interested in since that feeds into the calculation of kinetic energy for each cycle of bottle motion. It turns out that the kinetic energy in one impact of snow using the shaking method is about four and a half times the potential energy of the snow using the gravity method. In addition, the rate of impacts is about six times the rate for the gravity method. Thus, the shaking method is about 27 times as efficient as the gravity method. Still, it would take about nine hours to melt the snow using the shaking method, and that is assuming my arms don’t fall off before that point as a result of doing that much shaking. Nine hours is still too long.

Conclusions

In conclusion, none of the methods I tried turned out to be very effective for melting snow inside the insulated water bottle. The amount of time calculated for each method is shown in the table below. For comparison, I considered how long it would take the snow to melt if I had an uninsulated container sitting in room temperature (with heat transfer via convection of the surrounding air), using similar analysis to a previous post. It turned out that that method would actually result in a shorter time to melt the snow than any of the other methods.

So how did I end up melting the snow? The truth is, I never got the snow fully melted during the backpacking trip. When I found that the snow was not melting in spite of my best efforts, I decided to try another experiment. I decided to see how long it would take the snow to melt inside the bottle. Sure enough, when I arrived home approximately 24 hours after I put the snow in the bottle, there was still a small amount of snow remaining. So an insulated water bottle is pretty effective at keeping a drink cold. And if you are out backpacking and see some snow you want to use for drinking water, I recommend using a fire to melt it!

If you are interested in the calculations, see this spreadsheet:

Backpacking Snow Melting Calculations Download

Kinetic Energy of Human-propelled Objects

Introduction

My goal with this post, as with many of my future posts, is to fill a hole in the Internet. That hole is readily accessible information about the kinetic energy of human-propelled objects.

You see, when I watch sports, I often think about physics. I think about energy and forces and friction and velocity. I’m not the only one that thinks about physics in connection with sports. In fact, physics of sports is getting to be big business these days. With professional sports being a multi-billion dollar industry, any possible advantage in competition is worth money. Sometimes science is utilized to help generate competitive advantages in sports. And on the other hand, teachers of physics all around the world have used sports for years to attempt to entice unsuspecting students into being interested in science. I found a great variety of web pages related to the physics of sports while researching for this post.

At any rate, one day while thinking about physics and sports, I thought of a question: what is the object that can be propelled by a human that has the most kinetic energy? There are various objects in sports that are propelled in various ways (thrown/ hit/ kicked/ rolled). What is the method that produces the most kinetic energy? What is the object that, when propelled, has the most kinetic energy?

After having this question, I performed an Internet search to find the answer. I couldn’t find it. I performed various searches, but was never successful in finding the answer. So I determined I would need to calculate the answer myself.

The math to calculate kinetic energy is not difficult. The equation is simply kinetic energy equals half of the value of the mass multiplied by the velocity squared.

So it becomes a matter of simply finding all the values and plugging them in to solve for kinetic energy. I selected a variety of possible winners for objects in sports and plugged in the numbers. The results are shown in the table below, sorted by most kinetic energy on top to the least kinetic energy on the bottom.

Results

First, I will discuss the results, and then make some disclaimers and notes about the data included. I do not include my sources for all of the numbers because the information is fairly readily verifiable except in a few cases as noted below. The first note I must add for consideration in discussing the results is that the numbers shown in the table are mostly maximum values. For example, the fastest speed for a human sprinting is about 12.3 meters per second, which represents Usain Bolt at his peak speed in an Olympic sprint event. This is obviously a lot faster than the average person can run. I applied the same rule to the different objects to find the fastest speeds they can go. So the average person will not achieve similar results to those shown in the table, and maybe not even close. But the maximum results give us a way to compare the different objects.

In looking at the results, the top entry is for a human sprinting, which I don’t consider to be in the category of “human-propelled object”. Thus, that entry is shown only for comparison purposes. Maximum kinetic energy of a human sprinting is approximately eight times the highest kinetic energy of a human-propelled object. This makes sense because when running, a person can continue to add energy to the “object” (their own body) as they pick up speed. The energy doesn’t need to be added all in one short burst as with any propelled object.

The rest of the results in the table were initially surprising to me, and perhaps are surprising to you as well. If I had thought about it some more and done some rough numbers in my head first, the results would not have been surprising (but sometimes one must simply calculate first and think later!). Let me explain. In the kinetic energy equation, the velocity is multiplied by itself while the mass is not. Thus, it would appear that the velocity of an object will be the dominant factor in determining total kinetic energy.

Thus, I figured a golf ball, which I knew must get to some pretty impressive speeds, would yield a high value for kinetic energy. A baseball also gets to relatively high velocities. As you can see, however, these are at the bottom of the table. It is the shot (which is the name of the spherical object that is “put” in a shot put try), with a relatively low velocity, that wins for the propelled object with the most kinetic energy. This is because the shot has a mass that is about 160 times that of the golf ball. While the golf ball does reach speeds five times higher than the shot (resulting in the velocity factor of the kinetic energy being 25 times greater), the shot’s overwhelming mass compared to the golf ball makes a greater difference in yielding a high kinetic energy value of about 750 joules.

The other Olympic throwing sports come right behind the shot put, with the discus around 600 joules and the javelin around 400 joules. A bowling ball, the first real object in the kinetic energy rankings that the average American sporting enthusiast has access to, comes next with about 350 joules. However, the bowling ball spends most of its time limited to rolling along the ground. Along with the preceding objects, the bowling ball is simply propelled into a field of play with no opportunity for further interaction by another player.

Thus, the soccer ball is the first object in the list that is actually “in play” during a sporting competition. That is, a soccer player has the opportunity to absorb the full impact of the almost 300 joules of kinetic energy of the soccer ball. Good thing soccer balls are relatively soft and have a large area of impact!

An arrow, at about 250 joules, and a bullet (typical 22 caliber rifle) at about 100 joules are included in the table for comparison. Obviously, the kinetic energy of a bullet depends significantly upon the type of weapon and bullet (as shown in a table on this blog post, which also includes the kinetic energies of some of the objects included here). However, it is interesting to consider that a soccer ball at high speed can have almost three times the kinetic energy of a typical 22 caliber rifle bullet!

Interestingly, a football and a baseball, when thrown at their respective highest speeds, have approximately the same kinetic energy at about 150 joules. While a football has approximately three times the mass of the baseball, the baseball can be thrown at speeds almost 70% faster than a football, resulting in approximately the same kinetic energy.

Further Discussion and Notes

When I first put the table together, I figured I would need to have separate categories for objects that utilize an external tool to be propelled (such as a golf ball, which is normally propelled by means of a golf club) and those that do not require an external tool, since I figured it would give objects an unfair advantage to have a tool to propel them. However, with the final results, it is clear that the winner, the shot, is an object that does not require an external tool.

One interesting find I had while researching the maximum speed of a football was this article about physics in football with some gross physics errors. The article mentions that the work done by a football quarterback is the force multiplied by the distance, which is approximately correct; however, the distance used in the article is the total distance traveled by the football instead of the distance that should have been used, that traveled by the football during the quarterback’s arm motion. Thus, the work done on the football is calculated at 67,000 joules instead of about 150 joules as it should have been (incidentally, the value used for the force applied is also incorrect), as the work done on the football should approximately equal the kinetic energy of the football as it leaves the quarterback’s hand. This just goes to show you can’t trust everything you read about physics on the Internet.

For baseball, you might wonder if balls can be batted faster than they are pitched. The answer is yes. The maximum speed of a batted ball can get upwards of 120 miles per hour (compared to maximum pitch speed of 100 mph). The forces involved in the collision of baseball to bat can be pretty spectacular, as explained in an article from Popular Mechanics.

It was somewhat difficult to find the maximum speed of a shot put. I ended up just using the world record shot put distance and the fact that a typical shot put launch angle is 40 degrees to solve for shot put speed. There is a related physics homework problem (with fictional data), but the result is that the equation to solve for velocity is:

This yielded a velocity of 14.4 meters per second, which makes sense with a related article that solves for a female Olympic athlete’s shot put speed at 13.5 meters per second.

An athlete putting some serious kinetic energy into a shot (the round metal ball)!

One object I considered including was an atlatl, but I found it to be not significantly different from an arrow. I thought perhaps hunters before firearms were available would have found a way to get a lot of kinetic energy into an object for survival purposes, but it appears that maximum kinetic energy is not the only consideration in launching a projectile for hunting purposes.

Another object I considered including was a curling stone. However, the speeds of curling stones in competition are so low (two to three meters per second) that the kinetic energy ends up being pretty low as well. I suspect with the right conditions, an athlete could find a way to propel a curling stone at significantly higher rates of speed. Since a stone has a mass of about 18 kilograms (about 40 pounds), the kinetic energy could get pretty high.

I learned a few things about archery. Increasingly sophisticated compound bows have been consistently topping previous arrow speed records in recent years. Also, there is a lot of thought put into the mass of an arrow, and the experts have a lot to say about that. Thus, it is difficult to come up with a good “average” number for arrow mass, but I put what I thought was reasonable based on my research.

Another interesting tidbit I found was that the javelin was redesigned in 1986 to keep throwing distances down. The world record had gotten over 100 meters, which was unsafe for some competitions. I just found it interesting that the record distance was intentionally limited when the usual purpose of Olympic competition is to showcase maximum human abilities.

In calculating the kinetic energy of the various objects, I have neglected rotational kinetic energy. Some objects have a significant amount of spin on them, such as a golf ball with up to 2000 rpm and a baseball with up to 1800 rpm. I won’t go further into that realm in this post, but it may be something to explore in a future post.

Conclusion

So the winner is the shot of the shot put event! So if anyone ever tells you they will give you a dollar for each joule of kinetic energy you can produce by propelling some sports-related object with your own power (I know, this happens to me all the time, too), the shot put is a good choice. You might want to practice your form so you will be ready. And if that circumstance never happens, I hope you enjoyed reading this post and considering the energy of human-propelled objects anyway.

If you are interested, here is the spreadsheet for calculating the kinetic energy of the various objects. You can expand to include your own objects if you want. Let me know what you discover!

Kinetic Energy of Human-propelled Objects Download