A bad day for your ego is a great day for your soul. -Jillian MichaelsOne of the most popular exercises at the gym is the treadmill. And why wouldn't it be? Whether you're running or walking, it's a great way to get your heart rate up, get your body moving, and for many people, a great way to burn calories.
But however you use a treadmill, there's one extremely simple thing you can do to dramatically intensify your workout: incline it!
If you're an outdoor walker/runner, this is the equivalent of going uphill instead of over level ground. There are many physiological differences in walking along an incline versus on level ground, but what does physics have to say about it?
Normally, if you're on level ground (or a level treadmill), you stay at the same level in the Earth's gravitational field.
But if you walk uphill (or on an inclined treadmill), you not only need to move forward at whatever pace you were moving at, you also need to climb -- a little with every step -- out of the Earth's gravitational field!
The Earth's gravitational field is no slouch, either. I'm an 80 kg individual, and for me to raise my elevation by just 5.3 meters (about 17 feet) costs me 4,200 Joules of energy, also known as one food calorie.
Now, if I actually exercise, I burn significantly more than one calorie by raising myself those 5.3 meters. Why? The two most significant reasons are as follows:
- I am not a perfect engine. This means, in order for me to do 4,200 Joules of physical work, I need to burn about three times as much in food energy in order to get that much useful energy out. Alas, our bodies are inefficient in that manner.
- When you exercise and then stop, your body doesn't know that it's okay for your heart to slow down for quite some time. So spending an hour walking uphill will elevate my metabolic rate for a lot longer than an hour!
Let's make a helpful table. We'll just look at the total distance you travel (e.g., if you walk at three miles-per-hour for one hour, you go three miles), put in the incline, and see how much extra physical work you need to do!
| Distance (miles) | Distance (km) | Incline (degrees) | Extra Work (Calories) |
|---|---|---|---|
| 1.0 mi | 1.6 km | 1 degree | 5.3 Cals |
| 1.0 mi | 1.6 km | 3 degrees | 15.8 Cals |
| 1.0 mi | 1.6 km | 5 degrees | 26.3 Cals |
| 1.0 mi | 1.6 km | 10 degrees | 52.3 Cals |
| 2.0 mi | 3.2 km | 1 degree | 10.6 Cals |
| 2.0 mi | 3.2 km | 3 degrees | 30.6 Cals |
| 2.0 mi | 3.2 km | 5 degrees | 52.6 Cals |
| 2.0 mi | 3.2 km | 10 degrees | 104.6 Cals |
| 3.0 mi | 4.8 km | 1 degree | 15.9 Cals |
| 3.0 mi | 4.8 km | 3 degrees | 47.4 Cals |
| 3.0 mi | 4.8 km | 5 degrees | 78.9 Cals |
| 3.0 mi | 4.8 km | 10 degrees | 156.9 Cals |
| 5.0 mi | 8.0 km | 1 degree | 26.5 Cals |
| 5.0 mi | 8.0 km | 3 degrees | 79.0 Cals |
| 5.0 mi | 8.0 km | 5 degrees | 131.5 Cals |
| 5.0 mi | 8.0 km | 10 degrees | 261.5 Cals |
| 10 mi | 16 km | 1 degree | 53 Cals |
| 10 mi | 16 km | 3 degrees | 158 Cals |
| 10 mi | 16 km | 5 degrees | 263 Cals |
| 10 mi | 16 km | 10 degrees | 523 Cals |
This is all for a person with a mass of 80 kg (about 176 pounds). Isn't that a spectacular difference? In other words, if you make a long-term change from walking on a flat ground (or treadmill) to walking up inclined ground (or an inclined treadmill), you burn extra energy with every step you take!
And what's with the Jillian Michaels quote? Well, I'm no longer the fittest guy on scienceblogs; say hello to Travis and Peter over at Obesity Panacea, our newest ScienceBlog! But whatever you're doing, don't forget to take the time to get out there and do something active; you'll feel better and you'll be healthier. And who doesn't want a higher quality of life?
Physical Science
Medicine & Health
Comments
And this is why I bought a NordicTrack Incline Trainer X3. It goes up to 40% incline!
Posted by: Brian Shiro | March 10, 2010 8:41 PM
When you exercise on an inclined treadmill you do *not* climb out of the Earth's gravitational field, but you keep your distance from the center of the earth constant. So, what do you want to tell us here?!
Posted by: Duncan Ivry | March 10, 2010 8:43 PM
Duncan, every step you take is a step out of the Earth's gravitational field. It's not the Earth's fault that the treadmill is constantly pulling you back, and deeper into the Earth's gravitational field! (Stair-steppers work the same way, FYI.)
Posted by: Ethan Siegel | March 10, 2010 8:49 PM
Some physics isn't quite right here.
Last time I was on an inclined treadmill, my head (and attached torso) did NOT go "UP" as I hiked along. Most of me stayed at the same gravitational potential.
I did have to lift my legs more than on a flat treadmill but that, near as I can tell, was the only additional work I had to do. Not nearly as much work as climbing a hill.
Posted by: JDStackpole | March 10, 2010 9:22 PM
JDStackpole, I would think it is MORE work than climbing a hill because on a hill you get higher up and the gravitational field is weaker. On a treadmill, you stay at the same field level.
Posted by: NewEnglandBob
| March 10, 2010 9:33 PM
Nope. Gravity just doesn't drop off that fast. It's a minuscule (and undetectable by a climber) difference on any hilltop or mountain top relative to the valley.
Posted by: JDStackpole | March 10, 2010 9:53 PM
am a fan of your blog. will your next post be the physics of manny pacquiao's power and speed? i love boxing and wants to know a bit of physics too. :) thanks.
Posted by: bonvito | March 10, 2010 10:34 PM
No, really, I'm not lifting my body up any distance at all. It stays just about level even with a steep incline. When I climb actual stairs, my body goes up. The only extra work being done is lifting my legs a bit more. It's high-stepping; not a bad thing but not the same as climbing a hill.
Posted by: george.w | March 10, 2010 10:37 PM
the effect of altitude on gravity is relatively negligible. but if you wanted to get real exact... yes, you would be expending less energy climbing at an altitude of 10,000 feet.
Posted by: Sam Pennella | March 10, 2010 10:38 PM
If you want to put anything to do with a treadmill in terms of work, you get zero because there is zero displacement at the end when you get off. Actually moving up or forward doesn't get you the right answer in this case. You have to look at this incrementally because you don't get back the energy you burn as you "fall" back down. That being said, there is a great deal of evidence to suggest that running on a treadmill isn't as hard as pounding the pavement. Many marathon runners have found this out the hard way on race day.
Posted by: geopig | March 10, 2010 10:38 PM
it definitely isn't as hard as running on pavement. you have to push off with each step effectively using more muscle fibers in your toes and feet.
Posted by: Sam Pennella | March 10, 2010 11:22 PM
"it definitely isn't as hard as running on pavement."
And it definitely isn't as hard on your joints as running on pavement. Save your knees - use a tread mill - (or run on dirt)!
When I do the treadmill instead of outdoors exercise, I like to put the incline up to 15% (max on that machine). I feel like I'm using a lot more of the power in my legs - and I am. The treadmill is moving me backwards and downwards every second, and every second, I clime to maintain my position.
Posted by: Jackal | March 11, 2010 12:03 AM
Hello Ethan!
Thanks for the blog shout-out and the wonderful post.
Many of the participants that volunteer to exercise in our weight-loss studies, especially the older individuals, really prefer to keep the speed rather moderate but increase the incline. This ensures their heart-rate stays elevated, and they get a decent workout without having to run too briskly, thus reducing some of the impact on their joints. On the other hand, when doing maximal stress tests on our participants using a protocol which progressively increases the incline of the treadmill until they fatigue - I've seen a few cases where people stop the test due to severe cramping and discomfort in the lower legs. Thus, while walking up an incline may help burn a few extra calories - this needs to be considered against the potential reduction in exercise time due to cramping, or discomfort. From my experience, 1-3 degrees incline seems to be very doable for most people.
Posted by: Peter Janiszewski | March 11, 2010 12:07 AM
As part of my daily session at the gym, I do 3 km on the treadmill (if I'm lucky) at 6 degrees incline, so that means I'm doing only an extra 50 kcals of exercise. Damn!
Posted by: waynerobinson4
| March 11, 2010 12:07 AM
I'm agreeing with Duncan and JD on this one. Unfortunately inclining the treadmill isn't going to increase your calorie consumption as much as actually climbing a hill. The only effect the gravitation will have is on the legs having to be lifted higher, feel the burn :)
Posted by: William Paysinger | March 11, 2010 2:00 AM
running on and incline on a Running machine is not as hard as running on the road. The have to take out the effort afforded by the motor moving your legs backwards,
Posted by: John cornish | March 11, 2010 3:03 AM
According to Sir Isaac if a body has a relative motion in the vertical direction that is not accelerating then the laws of physics act as in every other non-accelerating frame including zero vertical motion.
While not actually rising, the legs fight the same acceleration of gravity as if the body were.
The same relative velocity effect acts in the horizontal so while your body remains on the treadmill you are pushing backwards the same as if walking the slope but without boundary of the machine.
Posted by: Ishmael | March 11, 2010 6:58 AM
Another advantage to incline training is that it can be less stressful to your joints. If you walk at a steep incline you can burn as much, if not more calories than if you are running on a level terrain. Running causes extreme impact. In fact, I read where running at a fast pace can cause the impact to be 2.5X your body weight. That's why many runners have to find alternative forms of exercise as they age. The popularity of elliptical trainers is partially from aging baby boomers who are looking for a form of exercise with less impact.
Posted by: Fred Waters | March 11, 2010 11:38 AM
While not a physicist (I used to be a chemist, but fortunately I have since fully recovered) I was at first surprised how many commenters don't agree that inclining the treadmill means working harder against gravity, even though the runner stays at the same elevation. But thinking about it, I can see why it's not immediately obvious.
By the same token, why does running on a treadmill on the flat require work (once up to constant speed)? After all, Newton's laws say that no force acts on a body that's not accelerating, and work done is the (integrated) product of force applied by distance travelled. The reason is because the runner needs to exert a backwards force along the bed of the treadmill to overcome its friction, which would otherwise pull him/her backwards. (Note that a somewhat greater force is needed when running without a treadmill, as then the runner moves relative to the air and has to overcome its resistance as well as that of the floor. Also, there must be some secondary forces needed to stabilise the body and avoid falling over.)
It's now much easier to see that if the treadmill bed is inclined, the force acting along its length has a gravitational as well as a frictional component, and the runner has to work harder to overcome both.
I know this is just a long-winded version of what others have said, but hopefully helps to clarify what's going on.
Posted by: Hilary PhD | March 11, 2010 11:46 AM
Me again. Just to add to what Fred Waters and Peter Janiszewski said, I don't think it's just about avoiding impact.
For a given type of exercise, you can raise your work-rate either by moving faster or by increasing the resistance to be overcome and therefore the force you apply. Some people's muscles seem to suit increased speed, others increased force. Modern cardio machines allow inclines and variable resistance so that each person can find a balance that suits them personally.
Posted by: Hilary PhD | March 11, 2010 12:06 PM
Ethan: "... It's not the Earth's fault that the treadmill is constantly pulling you back, and deeper into the Earth's gravitational field! ..."
No, it does not work like that. people on a treadmill stay at the same position vertically and horizontally all the time, and the surface of the treadmill roles away under their feet. If you want to have the effect of "pulling you back" you first have to run several steps forward and upward, then slow down, and let the treadmill *then* pull you back -- and this again and again. I have never observed people doing it like this.
It is like JDStackpole said: "I did have to lift my legs more than on a flat treadmill but that, near as I can tell, was the only additional work I had to do. Not nearly as much work as climbing a hill."
And above all that, exercising on an inclined treadmill is bad for the joints in the long run -- pun intended -- and even if Fred Waters above says the opposite -- because the feet are in a less natural position all the time. Because of this its also bad for the Achilles string (I hope this is the correct word). I would even say: If you want to ruin your feet, use an inclined treadmill.
To Hilary PhD:
A lot of physical work -- may be, even most of it -- has to be done inside the body. Different muscles have to move the body's parts in different directions again and again. Think about the heart, the lungs, the brain, etc., and the transport of all the substances the body needs.
Posted by: Duncan Ivry | March 11, 2010 12:53 PM
when you jog on a treadmill, you're primarily using your femoral extensors and flexors as well as the gastroc and anterior compartment. you don't use plantar flexors nearly as much as you do when running on pavement.
for an incline, you're adding a stair stepping motion to the run. => more calories.
that's the bottom line.
run for 3 mi on a tredmill one day. take a break. then 2 wks later, run 3 mi on pavement. I guarantee you that you will feel much more soreness in the dorsum of your feet.
add an incline to THAT... and there's your difference
Sam Pennella
Posted by: Sam Pennella | March 11, 2010 2:33 PM
@Duncan Ivry
I agree some internal work has to be done - I mentioned stabilising the body and Ethan mentioned physiological inefficiency (his indented point 1). Neither of that alters the fact that the force the runner has to exert against the treadmill is increased by a (quite significant) gravitational component if the treadmill is inclined.
Please learn something about the physics of vector forces.
Posted by: Hilary PhD | March 11, 2010 4:32 PM
Please review and update, if necessary, your contact information. I was unable to e-mail some relevant material to you that is too long for the comments.
Posted by: Bernard Leikind | March 11, 2010 5:31 PM
Hilary PhD: "... the force the runner has to exert against the treadmill is increased by a ... gravitational component if the treadmill is inclined."
I do not know, where this gravitational component should be. As far as I can tell, a person on an inclined treadmill has an inclined orientation of his/her feet, but nothing else.
"Please learn something about the physics of vector forces."
Thank you for pointing me into this direction. A poor, old mathematician has difficulties understanding just this ;-)
But to the point:
Let us observe a person on an inclined treadmill! Does the person move to a greater distance in the gravitational field of the earth, i.e. in a direction from the floor to the ceiling? No.
Is there work done by going along a path -- going along a path! -- up in the gravitational field excerting a force against gravitation pulling down? No.
We have to consider the "the line integral of F dot ds", with F = force, ds = infinitesimal path; see e.g. Richard Feynman's Lectures on Physics, page 14-1. Interestingly, Feynman writes on page 14-2: "... running upstairs is considered as doing work ..., but in simply holding an object in a fixed position, no work is done." Then Feynman goes on explaining how physiological work occurs -- muscles, nerves, etc. -- and how it differs from physical work.
The person on the treadmill is the "object", and the person holds "itself" in a fixed position relative to the gravitational field. Ergo: No physical work is done.
This is *the* important piece: No path, no work!
@ Ethan: I think, you have access to the Feynman Lectures. Please read these two pages, and tell what you think.
Anyway ... this is my last comment on this topic.
Posted by: Duncan Ivry | March 11, 2010 7:11 PM
But - your height is not changing at all although your footing may be more awkward. I'm with JDStackpole; I can't imagine the slight increase in work being anything like climbing a mountain.
@Brian: 40 degrees is extremely steep; I carry rope when I walk up/down terrain with that grade - do you really use it at such a setting?
Posted by: MadScientist | March 12, 2010 4:37 AM
@MadScientist he said 40%, not 40 degrees, so it's more like 22 degrees (I think)
Posted by: Emlyn | March 12, 2010 5:48 AM
Duncan Ivry asks: "Is there work done by going along a path -- going along a path"
The answer is yes. The path is the motion of the person RELATIVE to the bed of the treadmill.
I rather wonder how many of the people who talk about "slight increase in work" have any actual experience of walking quickly (I wouldn't run, but I hate running on the flat as well) on an inclined treadmill.
Posted by: Hilary PhD | March 12, 2010 5:58 AM
Skipping over the "I'm not going up on a stairclimber, therefore I'm not doing any work" silliness, there's another interesting issue here:
The "extra work" figure is only meaningful if you have a "baseline work" number to compare it to.
That makes for an interesting order-of-magnitude physics calculation. Obviously if we had wheels, we could "run" on a horizontal treadmill with (in the limit of perfect wheels) no effort. So where's the work coming from? I'd guess two main contributors:
1) The up and down motion involved in running and walking (in theory if our bodies were perfect springs, etc. we could regain the energy expended, but I don't think that's the case)
2) The acceleration of our legs and arms back and forth (again, if our muscles could act as springs, etc. we could regain this energy, but I'm pretty sure that's not the case).
Neglecting #2 and just considering #1, I'd guess that at each stride we're looking at an increase in body center-of-mass height that's roughly 5 to 10% of the stride length. So your baseline work would be equivalent to the "extra work" associated with the 5 to 10% incline numbers.
Posted by: Anonymous Coward | March 12, 2010 3:44 PM
Wow - very scientific - enjoyed reading your explanation. I always walk at an incline on my treadmill. I set it up to the maximum incline and work out -
Posted by: Sandy | March 12, 2010 6:39 PM
uh...as a physiologist you're a pretty good physicist. Your body is not stupidly fooled--metabolic rate (and, essentially, therefore heartrate) remain elevated post-exercise for good reason: there is a lot of chemical work to be done to mop up after exercise.
But the extra work in "high-stepping" is not all (or even mostly, I bet) being done by the leg that is stepping up--your power leg pushes the hip forward. On an incline there is an additional upward vector to that pushing force, and that accounts for most of the additional energy required.
I think.
Posted by: Sven DiMilo | March 13, 2010 11:41 AM
When walking horizontally, you place your front foot on the same level as the rear foot. Then walking up a treadmill at 10 degree elevation, and taking steps of 2 ft, the front foot is placed about 4 inches higher than the rear foot.
As you begin to transfer your weight to the front foot, your body must rise further than for a similar horizontal pace. The net effect is smaller than when walking up a 10 degree ramp because the treadmill surface is moving steadily backwards.
I suspect that an engineering approximation is that the body rises about 2 inches higher during each step, compared to the trajectory for walking horizontally. Calorie consumption would then perhaps be about the same as for walking up a ramp with half the treadmill slope.
Are there experimental measurements of oxygen consumption on treadmills of different slopes which could resolve the issue.
Posted by: Keith C | March 13, 2010 5:04 PM
You guys and your judgements, this is a great article and all I know as an experienced marathon runner, is when I do hill work on my mill, I sweat and feel pain, liek I have run 20 miles. When I run in outdoors and do hills, I sweat and feel pain. Incline running, walking or anything to my body is a workout and a half. Level running or walking isnt as intense as an hour of hills. Just go out or stay in and test it. Weigh yourself before and after on a week of hills and level running.
Posted by: Annette | March 17, 2010 12:03 PM
Having a master degree in physics, it is a bit frustrating to read people clearly not understanding the basic principles being so cock-sure. If you run on a treadmill, you are lifting your whole body not just raising your feet or whatever. The easiest way to realize that is to imagine that the treadmill is very long, say 10 miles with an inclination of 10%. Then if you are standing still at the top of the treadmill, the treadmill will carry you down 1 mile vertically after a while. If you on the other hand, run with the same speed as the treadmill your vertical position is unchanged. So who did the work to prevent you from "falling" the 1 mile? Your body and nothing else. QED.
Posted by: Richard | March 22, 2010 1:26 PM
Valuable information. Incline is pretty much common on treadmills now. It is hard to accurate differences between running on treadmill with incline and really uphill running. However incline is great feature for simulating uphill running.
Posted by: Treadmills Australia | March 30, 2010 1:05 AM
But the extra work in "high-stepping" is not all (or even mostly, I bet) being done by the leg that is stepping up--your power leg pushes the hip forward. On an incline there is an additional upward vector to that pushing force, and that accounts for most of the additional energy required.
Posted by: scan tool | June 10, 2010 8:05 AM
you're a pretty good physicist. Your body is not stupidly fooled--metabolic rate (and, essentially, therefore heartrate) remain elevated post-exercise for good reason: there is a lot of chemical work to be done to mop up after exercise.London
Posted by: London | June 15, 2010 4:29 AM
"Individuals undergoing SERE training are obviously in a very different situation from detainees undergoing interrogation; SERE trainees know it is part of a training program," Bradbury wrote, borrowing from the IG report's conclusion.
Posted by: tower defense | June 16, 2010 3:33 AM
I have a question for you Richard:
You said this: "So who did the work to prevent you from "falling" the 1 mile? Your body and nothing else."
Lets say you are standing next to a 1 mile high cliff. If you jump off, you will fall 1 mile. If you just stand on the ground at the top of the cliff, you don't fall. So who did the work to prevent you from falling the 1 mile here?
Posted by: Cameron | June 20, 2010 1:31 AM
Ms. Yoshida's still better than I ever was though; my top clocked pitch ever was only 62 and 58 was more typical on a very good day.
Posted by: Web Hosting Reseller | June 21, 2010 1:47 AM
One step at a time, here is my theory:
When you lift your left leg up and forward, you have to lift your legs higher than without the incline. This takes some work, but lifting your leg is just a fraction of your body weight.
Your right leg now is just sliding downward and your body overall just stays in place. No work. If you were climbing a hill, your right leg would now actually have to lift your entire body weight (against gravity).
Your body is say 4 times heavier than just your leg, so on a treadmill you only burn a fraction, ~1/4 of the additional calories that you would burn by climbing a real hill.
For example if running flat burns 1000 calories and running up a real hill burns 2000, you would burn about 1250 calories by using the incline.
Posted by: Johnny | June 24, 2010 7:08 PM
Very good points Johnny.
I have been thinking about this problem alot and have come to identical conclusions. The way I say it is that it depends on how you walk.
Consider stand the case where you stand still and let the treadmill pull you down, then instantaneously walk up (so the treadmill is basically immobile since you move so fast), then repeat. As you walk up, you have to spend just as much energy as you would on a hill, made obvious since the treadmill is not really moving. As it pulls you down, that energy gets put into the treadmill and motor system. So, if you were able to walk this way, then it would take as much energy as walking up a hill. Note that you have to provide energy both as physiological (or internal, whatever) work and physical work.
Consider however the other extreme: you somehow manage to stay put without moving your body at all. Say if you had roller blades on). Then you would be "walking" the treadmill without having to expend any energy. I probably need to at this point explain why you wouldn't roll down the treadmill, so I will. Let's tie a rope around your waist that fixes you to the frame of the treadmill. This rope stops you from sliding down the treadmill. The wheels stop you from having to move and keep your gravitational potential energy constant. I also need to point out (and I'm sure lots of people will argue this because they don' understand what work, energy and forces really are. I'm just going to ignore them. They can go read a physics textbook then come back) that the cord doesn't give you any energy. It doesn't do any work. Remember that work is force times distance, or more technically the integral of a force over the path parallel to its direction. Either way, the rope exerts a constant force on you, but since neither you nor the rope move at all, it doesn't have a distance to multiply by, or a path to integrate over. It therefore does no work. So by this mechanism, you "walk" the inclined treadmill without doing any work against gravity. In this example, you do neither physical nor physiological work. Note too that if you wanted an example where you did do physiological work, but no physical work, then just use a skateboard and do jumping jacks on it or something.
Now we go into a more realistic scenario, like the one you mention Johnny. Doing this, we have to make assumptions about how the person walks. For example: their entire torso remains entirely still, only their legs move. You can then say that they fit somewhere in between the two scenarios above. They don't have wheels, so don't do zero work, but their torso also doesn't move, meaning they don' do as much work as if they stood while going down, then walked all the way back up. My conclusion at this point is that an inclined treadmill takes more work that just high-stepping, but less work than an actual hill. I believe that should satisfy most people (as long as you agree with my arguments, if not, please let me know where you think I went wrong).
I would be interested in seeing some actual data that could be used to verify Johnny's theory and my arguments. I see the biggest challenge in to be how to measure how much work the human being does.
p.s. in case anyone is interested, I'm a third year electrical engineering student. I believe that even with years of education on physics and mechanics, you could incorrectly answer the question by applying logic incorrectly. I also believe that "common sense" can often produce the incorrect answer. Although not conclusive, or quantifiable, I value people with experience running bringing their opinions forwards. What I believe is truly required to answer questions like this one is a bit of physics understanding, a bit of experience in real life, a lot of thinking and logic, and even more discussion... thank God for the internet.
p.p.s. please bring forwards your opinions, especially if you disagree
Posted by: Cameron | June 24, 2010 10:24 PM
Cameron and Johnny,
This horse was dead and buried months ago; Ethan slew it definitively in his follow-up ("Einstein") post,
and dozens of people in both threads have added nails to the coffin.
But still the dissenters dissent. Sigh.
I think you can appreciate why few of us are willing to continue the conversation.
That said, suspending my own wariness for the moment, I'll assume you're sincere in trying to understand the physics and willing to work.
So let's cut to the core of the essential problem.
You're making it far too difficult.
Simplify the scenario as follows:
1) make the room a vacuum, to eliminate air resistance;
2) replace the human runner with an (idealized) electric car with 100% efficiency -- that is, all energy consumed goes into forward thrust;
3) the treadmill belt has constant velocity V, sloped at angle A from horizontal;
4) the car has constant mass M, and drives up the treadmill belt with constant velocity V, maintaining its position exactly.
How much energy will the car (i.e. its battery) consume in time T?
Hint: the answer, simple and unambiguous, will be in terms of V,A,M,T, and g, the local acceleration of gravity
(assume Earth sea level, where g=9.8 m/s2).
Once you find that answer and understand it, most of the complexities you've raised will be apparent as distractions.
Follow-up questions:
1) Where does the drained battery energy go?
2) Put the whole setup in an elevator moving up or down with constant velocity. What happens?
3) Now move the whole setup to Jupiter, or the moon. What happens?
4) Why is it appropriate to call this energy transfer "gravitational work"?
5) Why, when talking about gravitational potential energy, do we need to identify reference frames?
Posted by: ElmerPhuD | June 25, 2010 2:48 AM
This reminds me other first time I went hiking. The trail head said 1 mile...no problem I thought. 4 hours later I was up the mountain, longest and most grueling 1 mile I have ever walked. Never been on a treadmill...
Posted by: online fortune teller | July 9, 2010 6:11 PM
I have actually been to the other thread ElmerPhuD. I posted a comment on it in response to something you said.
I have spent a lot of time thinking about this problem. I also talked to other people I know and listened to their insight. Although I hadn't come to a definite conclusion before, I think I have settled with one opinion now - in some contradiction to what I thought earlier.
The approach I took is defining the person as a system and identifying the points where work can be done - i.e. energy transfer in or out of the system. I like simplifying the situation the way you did ElmerPhuD, with the remote controlled car. When you do this, it becomes apparent that the only way work can be done by the car is through the interface of the wheels on the tread. Since the car doesn't move up or down, no work can be done against gravity. So, the question is how much (if any) work is done here.
A quick free body diagram shows a force normal to the treadmill and another parallel - both to oppose the force of gravity. This parallel force could be a source of work, as long as we can identify a distance to multiply it by. And that is where I start to become less sure. Once argument could be that since the car doesn't move with respect to the earth, there is not distance - I don't like this. Another is that as the treadmill and wheels turn, the treadmill moves a certain distance over a time period with respect to the car. I believe this distance should be use. I would appreciate it if anyone had any insight into this.
By this assumption, a calculation of the work done by the car can be completed:
Parallel force = M*g*sin(A)
Energy = Force * Distance
Energy = M*g*sin(A)*V*T
1) All of this energy changes to heat through friction in the treadmill. Likely resulting in less electricity form the wall.
2) An elevator with constant velocity makes no change.
3) On a different planet (or accelerating elevator) the value of g (or net m*g with the additional kinetic "force") changes.
4) I do not think it is appropriate to call this energy work against gravity since no change in gravitational potential occurs. I can see how some people call it gravitational work since it results from the presence of gravity - either way, not a big deal what we call it.
5) Gravitational energy results from the interaction between two masses. It doesn't matter what reference frame we use, the change in distance between the two masses is what counts when determining the change in gravitational energy.
Someone I talked to brought up this hypothetical situation: assume the treadmill you are on is infinitely large (length and width). If so, to a person standing on it, it would seem like solid, stationary ground. Also assume that a uniform gravitational field is directed through the treadmill at a certain angle (without worrying about what has created this gravitational field). If the person stands still with respect to the treadmill, they loss gravitational energy and expend no energy. If they walk up the treadmill, it can intuitively be seen that they will have to do work. It is interesting to note that we can't exactly identify a reference point in the field of gravity and it doesn't matter how fast the person walks up the treadmill.
I am not sure about a person walking yet, because the situation is more complicated. This may or may not make a difference and I have yet to hear a good argument as to why it does or doesn't. For example; a person spends some time "in the air", and exerts force with one foot then the other.
Let me know what you think ElmerPhuD. I will be 100% convinced about the car once someone brings forwards a good explanation of the distance thing. I'm not yet convinced about a person.
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Cameron,
Kudos on the homework; you're almost there.
I'm glad to see you're sincere, so I'll try to explain in full.
Forget the runner for now; stick with the car until all is clear.
Main Question:
Correct. Energy = gMVT*sin(A).
From now on, let's talk about Power = Energy/T = gMV*sin(A).
That's the exertion rate for an ideal vehicle.
And it's the minimum average exertion rate for ANY vehicle/body maintaining position on an inclined treadmill.
Question 1) Correct: all such exertion goes into treadmill heat -- eventually. But it acts as *lift* first. See full explanation below.
Q2) Correct: additional components of vertical motion (e.g. elevator) don't change exertion. There's no such term in the equation.
Q3) Correct: moving to a location (e.g. Jupiter) with a different value for g changes exertion rate. I.e., gravity matters.
Q4-5) You're still confusing yourself.
First: accept the reality of Power >= gMV*sin(A).
Go run on a treadmill; you'll feel it.
Every interpretation of the problem, using forces, potential energy, etc, is just a different accounting trick to describe that reality. Any formulation which makes that exertion appear to vanish is wrong somehow.
Second: stop fixating on the relative height between Car and Earth. Yes, that measure of gravitational potential energy is constant. But it's irrelevant; it answers the wrong question.
The chronic dissenter's argument, often smugly asserted, goes:
"Vehicle never changes height relative to Earth, so there's no gravitational work."
But there the thinking stops, failing to consider these implications:
Implication #1) if planet-relative height doesn't change on Jupiter, gravity still shouldn't matter.
Implication #2) if height does change, as with the elevator, now gravity should change something.
But by now you know those are both WRONG! So don't trust that argument.
Third: let's not put semantics before physics.
I wouldn't quibble about whether to call inclined treadmill exertion "gravitational" or not, as long as our terminology motivates the correct predictions.
But if purists, in refusing to call it gravitational, deny the reality of gMV*sin(A) or make either wrong implication above, their language has failed.
So I'm calling it gravitational work. Because it's work, and there's a "g" in the equation, and it predicts what happens on Jupiter.
Fourth:
Potential energy, like distance, can be defined relative to any origin. So let go of Earth as an origin; follow Ethan's "Einstein" advice and consider relative perspectives.
Set aside your omniscience about the scenario and consider only what the Car "knows": it's on a surface, sloped to a local gravitational field. That surface may be long or short, moving or still, attached to a planet or not. That surface, plus any attached planets, elevators, etc, constitute a Thing.
Earth, if it exists, is just another moving part within Thing; we don't know and don't care how it moves.
Our concern is only with the energy transfer (i.e. Work) from Car to Thing.
Assign Thing a reference frame with origin at some arbitrary point P on the treadmill surface. P moves downward at speed V*sin(A) relative to Car (and relative to Earth, if we cared).
Measure all height and potential energy relative to P.
There are two concurrent components of motion, one rising and one sinking:
Motion 1) Car is climbing relative to Thing, converting internal chemical energy to gravitational potential energy, relative to P, at a rate of gMV*sin(A).
That's the work we care about, the energy transfer between systems, from Car to Thing.
Now, if we want to be omniscient and figure out where that energy goes within Thing, consider the second "sinking" process:
Motion 2) P is sinking relative to Earth, one of Thing's moving parts. Alternatively, Earth is "sinking" upward relative to P, "falling" within Thing at speed V*sin(A). Its fall is braked by friction, converting potential energy (again, relative to P) to heat at the same rate of gMV*sin(A).
But that's all internal to Thing; Car doesn't care.
Notice: turning off the treadmill stops 2) but not 1); Car keeps climbing relative to P, exerting the same power. Thing just stores it differently, as internal gravitational potential instead of heat. Car doesn't care.
In summary:
Motion 1: Climbing (Car-->Thing): Chemical --> Grav. Potential
Motion 2: Sinking (within Thing): Grav. Potential --> Heat
It's tempting to combine them, making gravity disappear:
1+2: Chemical --> Heat, no net motion
But these two components of gravitational potential, although the same magnitude, operate across different system boundaries: once Car-->Thing, once within Thing.
Gravity participates TWICE, not NEVER. But only once acts on Car and counts as work between systems.
The "distance problem", as you call it, is merely a failure to distinguish system boundaries.
Finally, as for the runner: he/she is just an inefficient vehicle whose momentary power output wavers around an average of gMV*sin(A)+Waste.
This is all to say: our host Ethan, Professor of Astrophysics, is indeed correct -- something for reflexive dissenters to keep in mind.
Posted by: ElmerPhuD | August 1, 2010 6:04 PM
You guys and your judgements, this is a great article and all I know as an experienced marathon runner, is when I do hill work on my mill, I sweat and feel pain, liek I have run 20 miles. When I run in outdoors and do hills, I sweat and feel pain. Incline running, walking or anything to my body is a workout and a half. Level running or walking isnt as intense as an hour of hills. Just go out or stay in and test it. Weigh yourself before and after on a week of hills and level running.
Posted by: ev dekorasyon | August 2, 2010 12:43 PM
I doubt anyone is reading this anymore, but I have a bachelor's degree in physics, and there is no way that an inclined treadmill requires one to burn as many calories as running up an actual hill. We can get at this without considering any equations. Consider: someone running up a hill with incline A and someone running on a treadmill with incline A (with A being some measure of the angle relative to flat ground: degrees, radians, whatever) should being doing pretty much the same motion. But then the person running on an actual hill gains elevation (gains energy) whereas the person running on the treadmill does not. Assuming that both people are running at the same speed relative to whatever it is they're running on -- treadmill surface or ground -- we really have to conclude that the person running up the hill is spending more energy than the person on the treadmill. In denying this, you are more or less claiming that gaining elevation does not require energy, which is a definite violation of the laws of physics. Please shoot me an email if you don't understand this. The person claiming to be a former chemist is misinformed.
Posted by: Stu | August 11, 2010 12:56 AM
But however you use a treadmill, there's one extremely simple thing you can do to dramatically intensify your workout: incline it!
Posted by: cheap sunglasses | August 11, 2010 1:28 PM
So spending an hour walking uphill will elevate my metabolic rate for a lot longer than an hour!
Posted by: Bingo Deposit Bonus | August 14, 2010 10:55 PM
The difference between treadmill incline running and hill running is that when "climbing" on a treadmill, your body above your legs does not go anywhere near as much distance (vertical) as if you were to climb a real hill. The reason why running up a real hill takes more energy is because your legs now have to push the rest of your body mass up the hill. On a treadmill, your legs only do work on the force of gravity, not your whole body. Simple as that. I was recently told by a sociology student that I was wrong about this. Apparently engineering physics is stupid.
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Posted by: baby shower supplies | August 27, 2010 11:32 PM
If you ride a bike on a treadmill neither your body nor your legs rise up or down any more than riding on a road.
First, ride a bike on a treadmill at a 10% incline for 5 miles.
Then, ride a bike up a hill at a 10% incline for 5 miles.
Which will take more effort? Why?
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Posted by: Credit Loans | September 9, 2010 7:01 AM
Ok, we are all currently on a giant treadmill. if we start at the equator and face east and don't move for 24 hours, we have moved the equivalent of the earth circumference without putting forth work. If we look at the distance it would be considered negative distance which would be due to the lack of 'keeping up' with the speed of the earth. Now if we were to be able to move forward at about 1000 miles/hour we are now doing the same thing as being on a treadmill. Would this NOT be work? would it NOT be the same as running around the earth if the earth was standing still? everything we do is in reference to other things and it doesn't change according to the size of those things.
and as for those who insist that the difference is in the fact that outside you have to pick your feet up more, next time you get on the treadmill try to stay on it without picking your feet up. ha ha!
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Posted by: beeg | November 20, 2010 2:19 AM
As a physics teacher who has had to try to grapple with the best strategies for clear exposition of such concepts for most of a lifetime, what always intrigues me most about this sort of interchange is not the mechanics per se, but the never ending confusion about the mechanics. Which is the invariable result, in my view, of how badly and inappropriately physics and maths is taught at school and at university.
Reading the comments shows in particular how much more difficult it is to apply correctly the widely accepted principles of classical mechanics than most folks realise - even those who have laboured away for three years at an undergrad physics degree and believe that they should now have this down pat.
First and foremost, here’s the Colenso definition of mechanical work:
“A wee bit of work is done by a system A on system B when system A pushes or pulls system B a wee displacement in the direction of system A’s push on, or pull of, system B.”
(Note that I have deliberately eschewed here the word “force” and the word “cause”.)
In real world systems, contrary to what is taught at school and even in undergrad mechanics courses, few systems move very far in a smooth, rectilinear (straight) line. Nor, cruelly, as if that were not already bad enough, do systems, or parts of systems, even follow the path of a smooth circle. Even relatively uncomplicated systems move along loci that need to be described accurately by the geometry of conics – hence ellipses (of which the circle is a special case), hyperbolae and parabolas.
To add insult to injury, complex, more “realistic”, systems are even harder to analyse than the aforesaid “uncomplicated” systems because complex systems tend to perturbate: the individual parts follow paths made up of sequences of bits of conics. Yes indeed, real life, complex systems are complex, tricky buggers to analyse.
So the baby-step mechanics that most folks learn at school and uni needs to be understood as a deliberately oversimplified introduction to what can turn rapidly into some quite tricky 4D vector and tensor maths - and like so much introductory mechanics a pretty misleading intro at that.
In real world complex systems, forces are not constant and they do not act in a nice, straight forward fashion through the centres of idealised bodies with no extensions, again as shown in the introductory mechanics text books for the sake of pedagogical simplification.
So, in real world systems, we have to analyse moments of inertia not mere inertias; angular momentum not rectilinear momentum; angular acceleration not rectilinear acceleration; torques not forces; not to mention that the magnitude and direction of the various accelerating parts are themselves constantly changing because moments of inertia and torques have a irritating habit of themselves not staying constant throughout the range of a system’s movements.
Indeed, the rectilinear classical mechanics we get taught at school is best understood as just a special case of angular mechanics where the displacement of the force from the centre of mass of the system on which the force acts happens to be zero – so there’s no torque, no angular acceleration. Nice and easy intro for the purpose of pedagogy that then leaves almost 100% of physics students confused until the end of their days.
So, back to the human body. Forget the introductory classical mechanics you struggled with at school or even at uni. The human body is a complex system of rotating systems, of which the moments of inertia and the torques acting thereupon at any moment are constantly changing and need subtle and very painstaking analysis to puzzle out. I’ve spent the last forty years trying to mathematically model the human body in motion and still haven’t got very far. (But then I always was a slow learner and worker.) And that’s after we recognise that all animal systems constantly do work even as they rest; more work as they sit; more work as they stand; more work once they start to move.
The key of course to all this as always is the diagram. Lots of carefully drawn diagrams in fact. How many times have I railed at my students: but where Perkins is your diagram, you wretched imbecile? Answer of course never, because I’ve never taught anyone called Perkins. But if I had, then the rhetoric would be apt.
So, as an exercise start by drawing a naked human walker lifting one leg to walk up one step. You will need to draw lots of intermediate steps before your walker has finally stepped up onto the one step. All the parts of the body: the toes, the arch of the foot, the heel, the lower leg, the knee, the thigh, the pelvis, the abdomen, the upper back and upper chest, the neck and the head will need to be drawn. And don’t forget all the parts of the arms. You will need to estimate the masses, moments of inertia, torques, initial average and final speeds of rotation of each part to estimate the work being done by different parts of the body on other parts. Part of the problem is that the parts of the whole body are counterbalancing as they accelerate in order to maintain balance.
When you have wasted the best part of the next month or so of your life on your detailed drawings, it will gradually dawn upon you that while the overall step-up of the person onto the next step has required of course that work has been done against the person’s overall weight (this involves gravity) this is only one tiny part of the whole work picture. Why? Because to accelerate one part of the body, even in the absence of the earth’s gravitational field, requires a torque to be applied to that part. So mechanical work is done. Because as we move, the mini-systems of our body are accelerated.
Which is why we do work, lots of work, even when we move quickly or slowly on the flat. Even when we skate on ice apparently hardly lifting our limbs at all. Yes, even when there’s a following wind and we skate on ice! Because we are constantly accelerating parts of our body in order to skate along and this therefore requires work.
But even when we stand still, or sit, or lie down, our muscles constantly do work. Only when we are dead does this process stop. Hence, only when you have worked out the work that all the parts of the step climber's body does even when he or she stands still; then worked out the extra work the parts of the stepper’s body do when they move in a horizontal plane; then worked out the extra work that we all do when we climb up one step, will you be intellectually at the place where you can see straightaway that running “in place” up an incline on a moving treadmill will certainly require far more work than running on the flat.
In fact, of course, exactly the same amount of work, if you ignore wind resistance, as running up a stationary, rubberised mat at the same incline at the same speed.
Posted by: Chris CD | December 13, 2010 12:41 PM
i actually started walking on the incline a year ago, only on 5incline. and one day i went in early and saw this girl with the incline on 15, i ask how are you doing that, she said, try it, and it will change yout life. so,for 30 mins. with a quick 60 sec rest between 15min. ive been doing the incline 4days a week incline interval of 15 and 5 switching every 3 minutes, with 30sec breaks at 5degrees incline. ive lost 40pd overall, 20 in the year. im looking to up my time here, maybe 45 min. when ever im done it says iveclimb 775 ft. does anyone know what that is equvalent to with thing, like a building or mountains, is this a great thing im doing, whats next.lol
Posted by: joe | April 16, 2011 11:29 PM
Can it be that you are all reading far too much into this treadmill incline problem? With no incline the motor does all the work driving the belt against friction. The more incline, the less work the motor does, and the more work you do. I might just go get my power meter and get some real energy readings.
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Posted by: Magenballon | August 16, 2011 4:23 PM
I wonder if it is a problem of work vs. energy, which I think are measured in the same units. A man flapping his wings and gaining altitude will add energy measured by the ascent into his load, but if he flaps his wings only hard enough to maintain elevation he does no work that is reckoned by PE, but he still might burns lots of energy to fight gravity.
Posted by: matt | December 10, 2011 12:32 PM
In the case of inclined treadmill vs. flat, users are expending the same KE, but the incline is undeniably harder, and perhaps the total energy there is the PE neded to recover from a dropped back elevation one would find himself at if he stood still relative to the belt. This would be the same PE one would gain in hill climbing at a forward speed equal to the belt speed of the treadmill climber. Both will use the same amount of energy to stand upright and resist gravity. But if there is an overall difference, it may be accounted for in my health club’s machines, which seem to understate vertical ascent. Perphaps the ascent readout it is the equivlanent ascent of hill climing, which they think requires more energy.
Posted by: matt | December 11, 2011 12:25 PM
I too as other posters have commented cannot believe that the education level has eroded to the point people cannot figure out simple physics.
First of all the whole moving in the gravitational field is confusing and incorrect for the point that is trying to be made. For simplicity's sake, gravity has the same acceleration on an object everywhere at all times. Toss a ball in the air. It moves upward and slows at the deceleration of gravity. Then stops and falls back with the same acceleration. At all times gravity is the same on the object.
Next, this is an issue of people getting confused by the relative motion of the climber and the treadmill frame/ground/surroundings being zero. If a row boat in a river is rowed with a force equal to the current pushing against it, it will not move relative to the shore. If the same rowing effort were put forth on still water the boat would move at the same rate the river was moving. Add some trigonometry and vectors and this is true in two dimensions as well.
Try this experiment if you believe that you do not have to pull your own weight up, against gravity, on an inclined treadmill just as you would a hill.
Straddle the treadmill belt with a foot on the platform on either side. Turn the incline up and set the belt to an extremely slow speed. Reach up with you foot (either foot) and take a stride as high up as you can on the incline. Put all of your weight onto your stepping foot and release the other foot from the platform and into the air. At this point your knee of your stepping foot should be bent with your foot high on the incline and your back foot is in the air. Now DO NOT straighten your knee, that would be you lifting yourself up against gravity. Just wait for the treadmill to pull you up because you do not need to lift yourself against gravity right? Keep waiting, keep waiting... Have you fallen off yet? Obviously the treadmill did not lift you, you continue to move down with the same speed as the belt. You cannot continue on and take another step until you have lifted your body up, against gravity, with your legs. Until you do that, your a$$ will continue to fall with the treadmill belt.
Posted by: brian | January 4, 2012 9:44 PM