Accuracy of GPS on Uphill and Downhill Sections

dev123

Hi all,

A thought came to me yesterday when I was out running hills.

Would it be correct to say that GPS measures the distance you travel in the horizontal direction and that it cannot take into account the vertical distance travelled by a runner?

So when running a steep uphill or downhill section it would appear to someone looking down directly from above, i.e. a satellite, that you had slowed down even though you had maintained your pace from the perfectly horizontal section.

An extreme example of this would be running to a vertical ladder and clambering up the ladder at the pace you approached it at. I would expect that the GPS watch would tell you that your pace was zero mile/min as you are travelling towards its line of site and not across it.

I do see an elevation reading from my Garmin but does the watch have the ability to correct for the distance travelled based on a combination of the horizontal distance travelled and the change in elevation?

Just wondering if this would contribute to GPS inaccuracy on extremely hilly courses such as mountain running.

Cheers,
Dev

sconhome

The software can calculate the vertical elevation over horizontal distance and calculate the climb as a factor of grade.

Using your ladder anecdote. If your ladder is 6m lying down on the ground it is still 6m leaning against a wall. You can calculate the length of the ladder (pretend you don't know) by measuring the distance the ladder is away from the bottom of the wall and the height the top of the ladder touches against the wall.

This is elevation over a horizontal distance and Pythagoras gave us a theorem to work out the length of the hypotenuse ie ladder length.

And the leaving cert starts Good luck all!!

dev123

Thanks Amphkingwest,

I was thinking that it would be a glaring ommision if the software was not able to take into account the change in grade.

I was kinda hoping that it didn't take into account the gradient as my pace dropped severly on the hills yesterday and I was hoping it was a software error and not my quads that were the problem

Krusty_Clown

Hi dev123, because the angle of the slope we would be running tend to be pretty negligible (5%-10%) the actual difference between the long side and the hypoteneuse is pretty marginal anyway.

Your pace will slow on uphill climbs, but it's not due to the additional distance of the long edge Vs hypotenuse, but rather because of the additional vertical motion required (carrying your body mass), and using different muscle groups*.

* Or at least that's my uninformed best guess

sconhome

Does the pull of gravity get stronger the higher you go? Like an elastic band being stretched?

Or is it just the effort increases as you lift your body weight further with each leg extension?

dev123

Hi Krusty,

The hills yesterday definitely looked and felt greater than 5 - 10% but I think you are right. You would expect to expend extra energy overcoming gravity in the vertical direction when you are climbing.

I suppose it would only become an issue as you start approaching 45deg hills where, if there wasn't the adjustment that Amphgkingwest describes, you could potentially have an error of just over 40%.

donothoponpop

amphkingwest wrote: »

Does the pull of gravity get stronger the higher you go? Like an elastic band being stretched?

Or is it just the effort increases as you lift your body weight further with each leg extension?

The force due to gravity (g) is a constant 9.81 meters per second per second. The energy required to move a body h meters above its starting point is mass x g x h. So the gravity remains constant, but your potential energy gets stronger, as h increases.

Krusty_Clown

dev123 wrote: »

Hi Krusty,

The hills yesterday definitely looked and felt greater than 5 - 10% but I think you are right. You would expect to expend extra energy overcoming gravity in the vertical direction when you are climbing.

I suppose it would only become an issue as you start approaching 45deg hills where, if there wasn't the adjustment that Amphgkingwest describes, you could potentially have an error of just over 40%.

Hi Dev123, the 'perception' is that you are running up a hill with an angle of 45'. The reality is that even the steepest of hill climbs wouldn't be more than 15'. My local hilly route is a 100m climb (400 feet) over the course of a kilometer on the road. Feels bloody steep. Angle? 5.74 degrees. Some of the lads did a race up Carrauntoohil at the weekend. Approximately 819m climb over the first 4.5kms. Bloody steep. Angle? 10.98 degrees. Perceptions are a wonderful thing.

dev123

donothoponpop wrote: »

The force due to gravity (g) is a constant 9.81 meters per second per second. The energy required to move a body h meters above its starting point is mass x g x h. So the gravity remains constant, but your potential energy gets stronger, as h increases.

And this potential energy is what will benefit us on the way back down, that is if you are coming back down. If not, you don't get any of the payback.

I would suspect that while we benefit from this potenial energy on the way back down we don't get out of it what we put in cos we are still buring energy to keep our legs moving on the way down. If we ran up a hill and base jumped we would definitely see the benefits!!!

Krusty_Clown wrote: »

Hi Dev123, the 'perception' is that you are running up a hill with an angle of 45'. The reality is that even the steepest of hill climbs wouldn't be more than 15'. My local hilly route is a 100m climb (400 feet) over the course of a kilometer on the road. Feels bloody steep. Angle? 5.74 degrees. Some of the lads did a race up Carrauntoohil at the weekend. Approximately 819m climb over the first 4.5kms. Bloody steep. Angle? 10.98 degrees. Perceptions are a wonderful thing.

Thanks for the clarification Krusty. For story sake when talking to people who don't run they will remain 45deg climbs

PhoenixParker

Krusty_Clown wrote: »

Hi Dev123, the 'perception' is that you are running up a hill with an angle of 45'. The reality is that even the steepest of hill climbs wouldn't be more than 15'. My local hilly route is a 100m climb (400 feet) over the course of a kilometer on the road. Feels bloody steep. Angle? 5.74 degrees. Some of the lads did a race up Carrauntoohil at the weekend. Approximately 819m climb over the first 4.5kms. Bloody steep. Angle? 10.98 degrees. Perceptions are a wonderful thing.

It's also worth remember that because the angles you're talking about are small, the horizontal distance you're running and the actual distance you're running are pretty much the same thing.

Taking this one as an example - 819m on the vertical, 4500m on the slope, you've travelled 4425m on the horizontal. Despite the 18% slope, the difference between vertical and slope distance is only 1.6%.
For a more normal slope of say 5%, the difference is only 6m over 4500m.

dev123

PhoenixParker wrote: »

It's also worth remember that because the angles you're talking about are small, the horizontal distance you're running and the actual distance you're running are pretty much the same thing.

Taking this one as an example - 819m on the vertical, 4500m on the slope, you've travelled 4425m on the horizontal. Despite the 18% slope, the difference between vertical and slope distance is only 1.6%.
For a more normal slope of say 5%, the difference is only 6m over 4500m.

And that 1.6% is probably well within the margin of error associated with GPS.

Was about to ask how does a GPD measure elevation but decided to have a quick look at Garmin Connect and found this:

• What are Elevation Corrections? Elevation Corrections cross reference the horizontal position (latitude/longitude) provided by the GPS with elevation data that has been acquired by professional surveys. When corrections to elevation data are made, each trackpoint of your activity now contains the elevation from the web service, not the elevation provided by your GPS device.
  Garmin Connect selectively applies corrections to depict a more realistic representation of your elevation experience. Activities recorded from devices without a barometric altimeter are enabled with Elevation Corrections by default. Alternatively, activities recorded by devices with a barometric altimeter generally contain accurate elevation data and therefore Elevation Corrections are disabled by default. For those users who are familiar with the MotionBased Gravity service, this is the same service.
  

Thanks for the replies folks

PhoenixParker

your elevation experience

snigger

dev123

dev123 wrote: »

your elevation experience.

PhoenixParker wrote: »

snigger

No laughing down the back!!!

T runner

donothoponpop wrote: »

Your pace will slow on uphill climbs, but it's not due to the additional distance of the long edge Vs hypotenuse, but rather because of the additional vertical motion required (carrying your body mass), and using different muscle groups*.

* Or at least that's my uninformed best guess

Its an interesting point. On shallow climbs you can use a similar running motion to road running. If the traction is good you can even lean back slightly and pull yourself up the hill.

That is why many elite flat runners who have developed strenght endurance for flat racing can use this endurance to climb.

At a certain gradient especially with poorer traction teh running style must change and more specific uphill strenght and endurance where the calves and thighs are contracted strongly over a long period of time must be used.

You often hear hill runners complain about faring poorly on hills in road races.

I wonder is it becuase they are using a specific hillrunning technique more suited to steeper longer climbs and are unnecessarily losing stride lenght and cadence.

cython

donothoponpop wrote: »

The force due to gravity (g) is a constant 9.81 meters per second per second. The energy required to move a body h meters above its starting point is mass x g x h. So the gravity remains constant, but your potential energy gets stronger, as h increases.

Not exactly. 9.81ms^-2 is an approximate typical value for acceleration due to gravity, usually accurate around sea level. However F = mg (special form of F = ma when considering acceleration as being that due to gravity) is actually a simplification of Newton's universal law of gravitation (image below because BBCode doesn't facilitate typing it!), where m1 becomes m, and g is an approximation for G*m2/r^2, m2 being the mass of the earth, and r being the distance between the bodies' centres of gravity


Basically put, this means that at altitude (when r increases), the force of gravity actually decreases (because of division by r^2), but it so happens that even at the top of Everest, this results in less than a .3% reduction in value of g, compared to being at sea level, so it's pretty insignificant!

Krusty_Clown

D'pop: you are so owned*.

* I assume, as I don't understand any of it.

donothoponpop

Krusty_Clown wrote: »

D'pop: you are so owned*.

* I assume, as I don't understand any of it.

Gauntlet across face. I welcome it. However, as cython mentions, there is a minute difference over a large vertical distance- certainly for the sake of our argument, the force due to gravity when running up a mountain, is constant.

Of greater significance, when discussing mathematical error, is your (KC) dismissal of shallow gradient when running uphill. Cython, perhaps you could break an ideal runner up an arbitrary angle theta into vector forces, and we'll see how far we can take (of) this?

sconhome

cython wrote: »

Not exactly. 9.81ms^-2 is an approximate typical value for acceleration due to gravity, usually accurate around sea level. However F = mg (special form of F = ma when considering acceleration as being that due to gravity) is actually a simplification of Newton's universal law of gravitation (image below because BBCode doesn't facilitate typing it!), where m1 becomes m, and g is an approximation for G*m2/r^2, m2 being the mass of the earth, and r being the distance between the bodies' centres of gravity


Basically put, this means that at altitude (when r increases), the force of gravity actually decreases (because of division by r^2), but it so happens that even at the top of Everest, this results in less than a .3% reduction in value of g, compared to being at sea level, so it's pretty insignificant!

Gonna end up getting this thread moved to another forum.

dev123

To aid the discussion and because I have been out sick for the last three days and I would like to take the opportunity to use my brain I have provided a link to a vector diagram.

Breaking a runner up into x and y components we have

Fx=max

and

Fy=may

When running at a constant speed in a perfectly horizontal direction and neglecting the fact that the runner bounces in the vertical direction during a stride:

Fx=0 (as there is no change in velocity)

Fy=mg (where g is gravity, which for the purpose of this we will assume is constant )

For a runner running a hill with an angle of theta and maintaining a constant speed the force vectors become:

Fx=m(gsin(theta)) as the gravitational force, g, now has a component in the direction of travel up the hill

Fy=m(gcos(theta))

Lets plug in some arbitrary values:

m = 80kg
theta = 10deg
g = 9.81m^2/s

Therefore the forces acting on the runner during the horizontal section equate to:

F = 80*9.81 = 784.8N

The forces on the incline are:

F = 80(9.81*sin(10)) + 80(9.81*cos(10)) = 909.2N

which is a difference of 124.4N or approximately 16%

I may be way of the mark but this is what I remember from college.

Hopefully someone can fill in the alcohol-induced blanks

T runner

dev123 wrote: »

To aid the discussion and because I have been out sick for the last three days and I would like to take the opportunity to use my brain I have provided a link to a vector diagram.

Breaking a runner up into x and y components we have

Fx=max

and

Fy=may

When running at a constant speed in a perfectly horizontal direction and neglecting the fact that the runner bounces in the vertical direction during a stride:

Fx=0 (as there is no change in velocity)

Fy=mg (where g is gravity, which for the purpose of this we will assume is constant )

For a runner running a hill with an angle of theta and maintaining a constant speed the force vectors become:

Fx=m(gsin(theta)) as the gravitational force, g, now has a component in the direction of travel up the hill

Fy=m(gcos(theta))

Lets plug in some arbitrary values:

m = 80kg
theta = 10deg
g = 9.81m^2/s

Therefore the forces acting on the runner during the horizontal section equate to:

F = 80*9.81 = 784.8N

The forces on the incline are:

F = 80(9.81*sin(10)) + 80(9.81*cos(10)) = 909.2N

which is a difference of 124.4N or approximately 16%

I may be way of the mark but this is what I remember from college.

Hopefully someone can fill in the alcohol-induced blanks

Interesting.

An test of the difference in pace on an uphill with excellent traction compared to a flat with excellent traction might yield blanks if it was greater than 16%. Attributable to poorer running economy uphill. As the slope becomes more uneven, steeper (and slippier), running economy becomes more important, relative to the extra forces due to the incline.

robinph

The reduction in the strength of gravity is negligible for us as long as we remain earth bound. The oxygen content of the air would be a factor on a very looooong up hill though, or if the race starts at a couple of thousand meters, and then goes up further.