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A buddy and I have been debating the caloric expenditures of running hills. After a little research it seems there is a reasonable discrepancy out there about just how much "work" is done when running up and down hills.
There is a an old-school formula out there that most treadmills and exercise calculators use that take your calories burned at a given pace and multiply that by the grade (or incline) to get the marginal increase or decrease in running hills. For example:
[kcals @ 5% incline @ 7mph] = [kcals @ 7mph] + [kcals @ 7mph] x 0.05
To me it seems this equation has a few shortcomings and assumptions:
- Assumes that the relationship between CALORIC EXPENDITURE and GRADE is LINEAR. So going from 20% grade to 25% grade is just as hard as going from 1% to 6% grade.
- Assumes that running downhill is easier than running uphill, and to the SAME DEGREE. So running downhill 15% is just as easy as running uphill 15% is hard. Weird double speak I know.
- BOTTOM LINE IS that if you run in a loop at a steady pace all your uphills will cancel with your downhills and you get no calorie calculation credit for any "hills" that are in the loop, no matter how big. I think most can agree that a hilly run is almost always SLOWER and harder than a flat run (we should get credit!)
What they found is very complex to interpret, however Figure 1 seems to make it most clear to the "rest of us":
- The increase in calories burned going uphill DOUBLED when the grade increased by only 20% (for example going from 10% to 30% grade energy expenditure went from 5 to 10). By the traditional calculation our calories burned would've only gone up by 20%, rather than 200%.
- When running downhill, energy expenditure seems to follow the traditional rule until we reach about a -17% grade. At this point it actually gets HARDER to run downhill than uphill.
So the net result of all this is that when running in hilly terrain we are burning far more calories than our old-school calculators are telling us, even if we run in a loop where the uphills "cancel" the downhills.
To go a step further I took a 1 mile loop near my house where I run up (and down) a big hill. Using the traditional calorie calculation I'd burn about 130 kcals (based on weight and steady pace). If I add in the corrections for hills from this study it bumps my calories up to 220 (about 70% more; note that doing this calculation was tough and used excel, I will spare you for now).
So the challenge is twofold: 1) repeat this study a few more times under varying conditions to validate it, and b) change the fitness calculator world to account for this.
A third challenge is underway and might just be patentable, so I will hold off on unveiling it for now.
Happy Hills
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30 comments:
If you're musing about calories burned on hills that's one thing, though I'm not sure it's really that important.
As long as the hill repeats are building power, then that's all that's important.
Even if running is more about calorie burning than specific training goals, the longer, slower runs will do the work for you, not the shorter, faster stuff.
Cheers;
http://james.wanless.info
I've become convinced that there are some routes around Atlanta that are entirely uphill! At least that's how they feel sometimes.
Vancouver, too! Though my weather is probably a tad worse for riding outside than yours right now.
I figured the following formula from efit routine software that I have.
For the same distance
Extra Slope Calories = (No slope calories * slope * 0.05)
OR
SC = NSC*S*1.05
SC: Slope Cal
NSC: No slope Cal
S: Slope
This also satisfies 20% grade double calories:
But I do not believe this. I can run at 8.0 MPH with 2.5% Grade (actually measured grade) on my treadmill for 18 minutes. This should be equal to 9.0 MPH on the flat. When I increase the slope to 7.0% I can run at 6.4 MPH for less than 12 minutes. That means I am working harder. But my calculation shows that this is equivalent to 8.65 MPH on the flat. I will encourage everybody to do the same experiment. Run at low slope on treadmill and see how long can you run. In your next workout, run at a larger slope and at a speed that you can run for the same amout of time like previous. My theory is that if you get same amout of tireness or your heart rate is same after same amout of time, you are working as hard and you are burning calories at rthe same rate. If we can have a few people repeat this, we can figure out a better formula. According to my own formula it should be like this
SC = NSC + NSC*S*1.065
This should be good for both speed and calories. So if you can run 10 MPH on the flat, try running at 6.6 MPH at slope 8 and see if you get same tireness. Make sure you actually measure the treadmill slope. Let me know if you want to know how to do that.
wow excellent post prinda. I have no doubt that running uphill is substantially more difficult than just as is predicted by the grade alone in a linear fashion. It seems right that about a 20% grade (quite steep) is 100% harder, maybe more, than flat.
Another question though is the downhill... My article I cited reported that while it is "easier" to run downhill, it is not nearly as easy as is predicted, and does not counter the UPHILL grade equivalent.
In other words, when I run a hilly loop from my house (and end up at the same elevation, ie the same uphill as downhill) that CANNOT be counted the same as flat land running. You do get some break by running back down the hill, but not nearly what was put into the uphill.
One additional reason you might get more tired than expected when running severe hills is that you are using different muscles, which may not be conditioned to the same extent.
Thanks for the post.
Doc26.2, your theory of using different muscles makes sense, but I have been training on my treadmill at 7% slope most of the time. That means my grade running muscles are well conditioned. I run 2.5% slope only occasionally. I guess I should experiment at smaller grades. As weather becomes warmer here in Detroit, I will see what speed can I maintain on the flat. My treadmill will not go below 2.5%.
I agree with you on the hilly loop. Let's say I go 100 m up 20% grade then 100 m down 20% grade. This should be equal to 200 m flat, but when I went 100 m up I already spent as many calories as 200 m flat (20% double theory). That means coming dowm 100 m 20% was like resting.
prinda, how "tired" you feel and what you're able to run aren't really the same thing as calories burned at all.
Unless you're doing serious endurance running (>10 miles) you aren't going to be depleting the glycogen in your body. This means your "tiredness" isn't the result of you burning a set number of calories but rathe its a measure of other factors.
Specifically its mostly a measure of lactic acid, your blood pH, and how hard you're working your cardiovascular system based on how quickly you're burning calories.
The relation to calories burned is not linear.
What's more, it will tremendously vary from one person to the next depending on a given person's VO2 max, cardiovascular health, and muscle makeup ("fast" vs "slow" twitch muscles).
i quit caring about caloric expenditure a long time ago. i weigh very little and burn virtually nothing even during very long runs, but i figure this is a benefit to my glycogen stores. everything makes a difference in difficulty, ie. is the slope strait up or does it canter, the duration of time and length of each incline, etc. nevertheless, i'm sure running hills takes more energy since i feel like my ass has been kicked after almost every run, but it sure beats the mental trauma of running those long, endless flats. there's nothing more exhausting than heading out for a 10 miler and being able to see the end of it.
What a brill set of posts, thanks
I am just starting running again and doing a most of my running cross country. My experience is that there is a non-linear relationship between hill steepness and energy consumption, I do run up short 30 degree hills and to get up them I have to considerably drop my speed (perhaps half as fast) so for what it is worth I can well believe the double calorie theory up steep inclines.
Thanks so much for this, the journal article is excellent. I live in San Francisco, CA, and actually do jog (or try to!) 30%+ grade hills, but was frustrated by online calculators telling me it burned the same amount of calories as running flat. And running down those hills is quite difficult as well, makes sense it would take more energy. Now I wonder how many calories a day I burn just walking around the city, as I live on a very steep hill. Hmm. I will have to send this link to other San Franciscans.
Running on an inclined treadmill is not the same as running up a hill, although they both recruit the same muscles. The calorie expenditure on an inclined treadmill is due to the change in stride/movement. It may be that the movement is less efficient until you build the appropriate muscles.
The potential energy gained by running up a hill is never gained on a treadmill.
When running downhill, you are able to capitalize on the potential energy up to the point you are resisting the downward pull of the hill. The more you can let your legs 'Roll' the more of the energy expended climbing the hill is recovered. Once the downhill gradient is to steep, then it takes a lot of energy to regulate your downhill speed.
I am an engineer, and have pondered this since getting the record for my sports clubs "Elevation Challenge" which the highest "elevation gained" at 30 degrees for 5 minutes on a treadmill.
Unfortunately there is no elevation gained, and my performance up a real hill would be a lot less. I just won't tell anyone at the club yet.
The real calculation would have to be empirically derived, and would absolutely have to take into account the weight of the runner. It would also vary with the efficiency and fitness of the muscle groups recruited.
I can't really add much, several of the comments did a great job of getting more "scientific" ........but I will say this...... I work near rock creek park in DC. The park is very hilly, with some uphill/downhill grades of up to 30% for a couple hundred feet at a time. Since I started running these, and gave up on the flat road/treadmill route, I have lost considerable weight. So while my diet has stayed the same, and I am running the same distance that I was on the flats, I have lost a whole lot more weight. I do a loop, so zero net. Sometimes over 7 miles my phone (I map the routes with gps) will say 1500 net upward vertical (my run today). There is no way that those old school calculators can be correct. based on my pace and distance, an online calculator claims I burned 750 calories today (this was a flat calculator, but would apply to the net=zero calculator method). This is definitely false. I run a lot, and what I did today would leave me feeling like I would when I run almost double the distance at that pace on the flats. And like I said, I can't get very scientific, but I suspect I burned between 1000-1250 calories today. I would like to see a built in calculator on my phone that will calculate the calories burned based on the route that it records and some of the logic that people allude to in these comments. in the meantime, I'll just keep using my intuition to estimate.
I too have been frustrated by the lack of good calorie estimators for uphill running - brought home to me recently when I ran up a 300 m hill only to be told by my Garmin Forerunner 305 that the steep section, where I slowed to a crawl, had burned the fewest calories.
As a physicist I knew that was wrong, and I did a simple estimate: work done to move a mass m vertically through height h is mxgxh
This means that a 100 kg person climbing a 300 m hill increases his potential energy by approximately 300 kJ. If we assume the body is about 25% efficient, 1 kJ of work requires about 1 kCal of energy expended (convenient if slightly inaccurate conversion). So my climb would have added around 300 kCal to my expenditure - which is more what it felt like.
Interestingly, the Minetti paper had a graph (fig 2) for energy expended when climbing which had a lot of values in the range of 40 J/kg /mvert, which corresponds to 40 Joules to move 1 kg vertically by 1 meter. This is almost exactly the factor I got from first principles. In fact their graph 3 shows the "mechanical efficiency" for climbing is around 25% - again the factor I had picked from the air.
So I think that's a good rule of thumb to use - and I propose modifying the calorie expenditure equation [for climbing only!] as:
(kilo) calories expended in climb = calories on flat + (body mass in kg) x (vertical climb in m) / 100
Since grade = vertical climb / distance, we can also use
extra calories = (body mass in kg) x grade (in %) x distance (in km) / 10
Using my example above, grade =15%, distance covered = 2 km, body mass still 100 kg, you again get
extra calories = 100(kg) x 15(%) x 2(km) / 10 = 300
Round numbers - but I think they would be good enough for estimating calories burned (better than doing nothing, anyway).
The extra effort it takes to go downhill (our experience and the Minetti paper agree it's not a zero-sum game) is small compared to the uphill effort - I suggest you use "calories on the flat" .
Happy running all - and thanks for this interesting thread!
Wow that is the best things about running, burning calories.
I get very frustrated with the way my Garmin 405 ignores my extra efforts in the hills calorie-wise. I used to have one of the older Garmins which did seem to take elevation into account and I used it as a way of guesstimating what length a flat run equivalent would be which is handy for training. They obviously changed the formula however.
I did a 19 mile race with 1500m of climb yesterday and the Garmin awards me slightly less than 1900 calories for the effort. I tried Dr FPM's formula which gives me in the region of 28,000 calories which I reckon will be much nearer the truth as it took me 4 hours 20 mins and a marathon would take me substantially less time than this.
Thanks for the post! That explains why I've had 6 small snacks today and am still hungry!
I like your thinking Dr FPM, and especially that you did it for me. I want you to check my work though: I run trails with an average 15% grade and always make a loop. Does this mean that I can use your extra calorie equation and use half the distance?
Justin, I think that's right - if you are running a loop with an average 15% grade throughout, it is approximately equivalent to running half the loop up a 15% grade, and the other half running down again. So using my approximation (that only uphill running burns extra calories, but that you don't "get them back" on the downhill) you can use the same equation as before with half the distance of the total loop.
Good lick with that, and happy trails!
Floris
That should have been "good luck" obviously - %#>€+£|¥£ iPad keyboard. Spell checker obviously thought that was an OK thing to write. Ah, technology...
The thing about running on a treadmill with a grade though is you are just changing the way your feet land, your not acutally pushing yourself up a hill right?
Regarding the comment about "running on a treadmill is not the same thing as you do not gain potential energy" - I have to disagree. As you know, laws of physics are unchanged in inertial frames of reference. If you imagine your treadmill to have a really long belt, and your frame of reference (your "camera") moves at the same speed as the belt, then you will see the runner is really running uphill - and the forces on his/her legs are indistinguishable from those you get with a real hill. The only difference is in the air friction, which is worth about 20 sec/mile, and which is why most treadmills won't go below 2 degrees (roughly equivalent to running on the flat with no wind).
Bring on the counterarguments!
Thanks for this post - calories on hills has bothered me for a long time - both when trying to loose weight and train, and when wanting to ensure I am eating enough to fuel my training. I have worked out my routes with Dr FPM's formula - once I got my head round which units were to be used where and it feels about right - though now I am confused about where speed fits in. If I walk a hill then this forumla will give me the same extra calories as if I run up it. So am I really using the same calories thrashing up a hill as fast as I can in training (not all that fast actually) as if I saunter up it on a Sun afternoon walk - doesn't feel quite the same to me. My calory logging program has running calories by the min mile - does this come into this here - or is that what the 'add the flat running' figure deals with? Does my question make sense?
Emma,
It is always a good idea to check math against your gut. You asked "So am I really using the same calories thrashing up a hill as fast as I can in training (not all that fast actually) as if I saunter up it on a Sun afternoon walk - doesn't feel quite the same to me". The short answer is "yes - sort of."
When you look at calories burned when running a mile on the flat at different pace, you find there is very little difference in total calories, although the instantaneous effort is much larger with a faster run.
My formula looks at the basic physics of lifting an object - the (useful) work done is equal to the increase in potential energy, which is m x g x h . When you run faster you burn the same calories in less time, but it is almost the same number of calories (to what physicists call a "first order approximation". What they mean by that is "if you measure this carefully you will find I am wrong; but it is better than nothing")
I hope that helps?
Thanks Dr FPM that makes sense and has cleared up a query I have had for a while really. Thank you I understand :D
THANK YOU! Most my outdoor runs are hills! After running my first 1/2 marathon I learned some tough lessons regarding refueling. I did not realize how many calories I was actually burning during training until my clothes no longer fit and people thought I wasn't eating. I am more fit than I ever have been but I need articles like this so I can continue to learn.
i got the impression that the reason a treadmill incline doesn't give the same resistance as a hill is that the belt moves back whether you push hard and, thus actually move your body upward from its initial position or not.
i recently read "the first 20 minutes" by gretchen reynolds, and i'm pretty sure it was in there that i read that it definitely is not equivalent. if it wasn't in that book, i am fairly certain i read it somewhere equally reputable. you know, the result wasn't based on guessing why it ought or or not to be the same but, rather, on empirical measurement.
anyway, i feel pretty confident stating that much is not actually up for debate. real hills require greater effort.
You are right the the "slop" in the belt (the "give" when you push hard) will change the peak load on your muscles - making the treadmill more forgiving and a little bit easier on the muscles. From the perspective of a raw energy calculation this would probably translate into the muscles becoming a little bit less efficient - so you would burn slightly more calories if the belt was infinitely stiff. I suspect the impact on calories is smaller than the impact on muscle fatigue / training effect. As I said above - there is a measurable impact of air drag that makes a treadmill "lighter" than a real run at the same incline- you need to add about 2% slope to break even.
This thread is ancient, but I wanted to add something to the discussion.
From all the comments I have read about hills on various forums, everyone seems fixated on calculating the energy burned, but no ones ever brings up the POWER needed instead. I think discussing power is more important since from my own experiences, the body requires disproportionately more power to go up a hill.
In other words, the human body is not a machine and does not work in a linear fashion when running faster or up a hill.
Cool post. For those reckoning calories from a grade as (mass * g * vertical), as the work done -- don't forget the thermodynamic efficiency. Typ. 25% ... So you need to burn 100 kJ to get 25kJ of vertical potential energy. This also explains why you don't "get it back" downhill.
While I was with Dr FPM in the earlier post about elevation change (mgh) I have to disagree with this reference frame idea making inclined treadmills the same as hills. The potential energy effect depends only on the distance between the centers of mass of the two objects, and the distance between me and the center of the earth changed on a hill but not on a treadmill. Part of my body, from knee to foot does change potential each step up and down, as does part of my thigh, and maybe my torso goes up and down a small portion of the inclined step, but the mass of my body is not raised by the stride length times incline percent. Inclined treadmill certainly burns more than flat but it doesn't earn the potential energy term introduced above.
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