reading and writing on UL trips

[Roger B]

Well done Roman, this question seems to be a model for a recent unsupported long distance hike.

[Woubeir (from Europe)]

It haven't solved these equations in perhaps 7 or 8 years, but my initial calculation gave this as a result:
s(t)=14t + t²/2 with s(24)=624.

[Peter McDonough]

Toms right:

Here's some further explanation

Its first order separable, linear equation that does exist and have a unique solution.

First we move the dt to the right side, which gives us

ds = 14dt + tdt

Then you take the integral of each term, which gives us

s = 14t + 1/2t^2 + c

The constant of integration was left out by Tom, but since it comes out to be zero izzz all good.

Plug in the initial condition.

0 = 14(0) + 1/2(0)^2 + C

C = 0

s = 14t + 1/2t^2

s = 14(24)+1/2(24)^2
s = 336 + 288

s = 624 miles

Come on Roman is that all you've got?? Send me one that is at least a second order linear that only exists around a regular singlular point….ok??

:D

[Roger B]

Yes 624, but what is the connection between 624 and 24 and Roman?

624 was the easy part.

[Roman Dial]

You guys are great. I love this site.

24 days and 624 miles were the end points of my experience on the Arctic 1000.

While I had hoped to eat ~2lbs/day and improve our distance by ~2 miles/day, as I'd found from adventure racing, it actually came out to reduce by 2 lbs/day meant that we could go 1 mile more per day.

Possibly this latter result was because we made 12 hour days the norm rather than 18+ hours as in adventure racing.

And Crazy Pete: very well done…but know I have yet to come up with a good second order DE that we might feel pertains to us….but I will work on it!

[John S.]

After Ryan mentioned taking mechanical pencils in another thread, I found some mini mechanical pencils at a Walmart Neighborhood grocery store. They are in a pack of 5 pencils, just under 4 inches and are Foohy brand…for less than 2 bucks.

[b d]

I got the following after first Romanizing the equation. Starting with the original Romanesque Equation I first got:

DS/DT = XIV + T

Divide and multiply by D/D, then (S/T = XIV + T) BY D/D.

Assuming D is not 0, then D/D = I. Then I times (XIV + T) = ITSELF or UNITY. In other words it is a unified disintegrated differential equations.

Then S/T = XIV + T OR T = (XIV + T)/S, if you divide both sides of the equation by S.

Thus, s(0) = O could not be true unless T = 0. So the only possible answer is that given by other scholars and trekkers who would know what the starting factors, location, and variables were (which appear to be Roman's appetite and food budget for trip in the Arctic). One would, therefore, make the assumption that the range of probable, if not possible answers would include the farthest distances one could go and the least simultaneously — fulfilling the requirement of diminishing simultaneous returns in closed dietary pack systems, not otherwise providing for the possibility of negative distances or 0 — requiring that there be a positiev but uncertain resolution to the equation.

If the equation were to be subjected to further testing and field experience, possibly deserving an article here or in the print magazine, then the numerical probability would probably be more certain and one could then hypothesize both an absolute maximum, the mean, median, and even negative limitation points — such as those sugggested by Pondering in the unpublished treatise "Negative Distance Effects of Hunger: The Simultaneous Distance Quandry of Positive Distance Traveled Total Miles Trekked Diminishing In Returning for More Food Events."

Also, as a secondary numeric anomaly, since T is iself equal to itself, a period of Time spent hiking for 24 days, with s(0) = 0, can only lead to a conclusion that d = d during the particular time as empirically measured and not deduced.

Therefore, the concept or mathematical idea that s or t leads to a change in location (d) is only a theoretical possibility in a range of locations from 0 to 624, based upon actual calculations and observations by trekkers in this time-space reference frame.

The ultimate conclusion: 624 is the best available solution, but is subject to human error and will require both verification by the BPL Staff who should weigh in with a study with thermometers, calibrated walking devices, and photographic evidence. The more likely conclusion, based on this equation is that nobody is getting anywhere fast. Even that is questionable within the ranges of possible, though not probable sets of whole integers resolving the equation for s(t) at the conclusion of 24 days.

[Peter McDonough]

BD– I am not exactly sure what you are doing, and I believe that you have made some simple mathmatical errors. First off, what exactly are you doing multiplying/dividing by D/D?? Are you just saying that the Ds cancel each other out because they are superimposed over a fraction line?? Because those are demarcations of the process of derivation, not variables, and to multiply by a D is incorrect, unless you are somehow intergrating somewhere else in the equation.

Secondly "S/T = XIV + T OR T = (XIV + T)/S" is NOT true, because when solving for T one would get

S = T *(XIV + T)
S/(XIV + T) = T

Thus your statement T = (XIV + T)/S is wrong.

And if you were actually solving for T in that equation you would factor it out, otherwise the entire excercise would be fruitless.

Thirdly, a solution DOES exist thanks to the Existence and Uniqueness Theorem, and a quick look at the graph shows it to be continuous in the examined area. I am thus confused at your statement that 624 is the best available solution, because it IS the solution for the IVP with given input value.

[b d]

Crazy Pete,

Oh. Oops. Sorry:) (Do you want help off that rock? Great avatar, by the way.)

I thought the processes or process of derivation was a constant, but the variability over the set of possible processes might itself vary, although not in this case — given that it is apparently a function or process of Roman's appetite, or any other trekkers or the multiple variabilities of either other trekkers or sets of trekkers dietary requirements or choices. Kind of like econometric regression models of consumer behaviors in the case of scarcity models based on multiple variabilities and differential processes. (*See below comment on the Existence and Uniqueness Theorum.)

Thus the overall apparent aberations from mathematical norms and rules. Also, while the use of S may appear to dictate your version of the equation it is also possible that speed will dimish the dt portion of the equation IMO, thus I chose the diminutive form of dividing by S to find the solution over any given range of T.

Sorry bout that. I still firmly believe that best choice of an answer (your solution) is 624, based on the previously described simultaneous returns quandary. But, I am willing to go along with any solution so long as I do not think about this when trying to go to sleep tonite.

* Re the Existence and Uniqueness Theroum from the University of Texas at the Permian Basin: "There are two important points to make now. First the theorem does not tell us anything about how to find a solution. Finally, we have no idea of how small the region may have to be to accommodate the theorem. In other words the interval of existence for the solution may include little more than the initial value." at http://www.utpb.edu/scimath/wkfield/mod3/Exuni.htm

[Peter McDonough]

BD

---

"Then S/T = XIV + T OR T = (XIV + T)/S, if you divide both sides of the equation by S.

Thus the overall apparent aberations from mathematical norms and rules. Also, while the use of S may appear to dictate your version of the equation it is also possible that speed will dimish the dt portion of the equation IMO, thus I chose the diminutive form of dividing by S to find the solution over any given range of T. "

---

End BD

S/T = XIV + T
(S/T)/S = (XIV + T)/S

1/T = (XIV+T)/S
Not T = (XIV + T)/S

T = (-14 plusminus (196-4s)^.5)/2

But this is all subiderary to the fact that what the equation states is that the rate position is changing with respect to the change in time (velocity), is equal to 14 plus t. Which means that the velocity is changing, which is the whole point of a variable slope. The velocity increases as T increases—FOR ROMAN. Whether or not the equation applies to others has no effect on the correctness of the postulate. Say for someone sitting all the time, the equation would be DS/DT = K, where K would be a constant velocity, in this case zero.

As to the E and U Theorem, it does not tell us how to solve for a solution. It gives whether one exists.

The graph of the equation proves continuity:

The interval of existance (-infinity, infinity).

Perhaps I've never learned what you are doing with the derivative–but treating the D as a variable automatically removes it from the equation, just as 3/3 = 1. D/D would be one as well, except it means the first derivative of position with respect to its corresponding first order time. So in essence, I am still confused by what you did…

And waaaaaaiiit a minute— time is the INDEPENDANT variable here. You don't calculate time based on distance, but rather distance based upon time.

[Roman Dial]

Crazy Pete,

How about you? Have you done any UL modeling you can share with us?

[b d]

Crazy Pete,

Quite right.

So if T is independent, then the issue becomes is the 14 correct if the equation is to work for the generalized equation for a trek of many people or individuals or people sitting at home.

One question, how do you get to:

T = (-14 plusminus (196-4s)^.5)/2

That may be my problem, even though this started as a kind of fun spoof thing for me, I take seriously what you are saying because my food could depend on it or people's safety.

I think the issue is, mathematically, you cannot include enough processes, differentials, or variables in different levels of mathematical process to make a formula that will work in this example. Although the usefulness of them is that they will prevent disasters by predicting a range of values which will work.

Again, while I was spoofing at first, thus the reference to the made up unpublished treatise, you guys are right on … I just don't think the "theorum" or "formula" or "equation" will work in realtity.

Thus, like Godels proof, even within a mathematical system let alone when it is applied to realtiy, the system cannot generate its own proof, or in the case of the general theorum of existence of unique solutions, this may be a kind of tautology. Input in = input out, or confirmation that given the intitial hypothesis there will only be one answer, regardless of the realities the formula is trying to measure. Does that make sense to you?

This is starting to interest me so much I can't stop thinking about it, but no harm meant on my side. I wrote a spoofy answer and got back what is expected at this site, a really great answer.

Update: T may not be independent, because of the length of time travelled each day, or the length of time it takes to travel x distance, or some other function of time, even though it is a constant as we measure it. Thus, c becomes important in your equation, because it may not be zeroed out because s(0)=0. So the question that really starts to loom, and I was afraid of this given my mind and interest in this, is: if c, in your equation cannot be established without empirical data, can it be cancelled out … and can s(0)=0 for more than one set of facts????

[Peter McDonough]

Roman– Not really theoretical stuff like this would be, but simply stuff like force on tarp panels in various wind speeds.

BD – I used the quadratic formula to solve for T.

The C is the constant of integration; briefly, that means it is similar to a y-intercept in that it moves a line up and down. We graph velocity vs time, where the slope is acceleration to determine total distance traveled by area under the curve. By the initial given condition, that S(0)= 0, C = 0. All this initial condition means is that the 0 distance has been traveled at time 0. If you start at say 5 miles already having been traveled, then S(0) = 5 and C would be a value other than 0 due to the face there is more area under the curve.

The great thing about the integral is that it takes into account that the distance traveled each day can be different, as hmmm let me make a graph.

OK, in the graph you can see that the velocity is constantly changing, just like you said. As such, a different distance is traveled everyday, but this can still be accounted for. The hard part is getting the equation for this line, which Roman provided. Everything else is (somewhat)simple mathmatics.

That said, I would not be able to use this for any acutally applications on reality due to the fact I do not have a formula for DS/DT.

[b d]

Crazy Pete,

Cool. Trying to put math around this UL stuff is cool, way cool.

I never thought of doing it. But I only have myself for a sample/null set v. sample. Thus you guys are way ahead of me on that one, and much more.

I think the problem is your null set or model will require, mathwise, some 15 people to make the N (or in your equation ultimately "c" valid). Then the "C" or "c" and the 14 will work in combo.

Then you can develope a functional equation that is correlated to reality for specific real groups or individuals. (Given the caveat that it won't work on any given day — the curve margins.)

My first thought is what needs to be done to make it work, to establish a curve, like Roman has started or suggested, you need to create a model of each function: DS and DT. (Thus your comment that DS/DT does not have a formula or reality at present — to me that means empirical data + thought.)

I believe it can be done and that is what this site is for, we just have to control for or establish what the ranges are for individual people or sets of people, trekkers, or whatevers that the DS/DT will apply to.

Then they will have to hold to a particular projection or model of behaviour to make the math fit the reality???

Edit: That is why I zeroed out the d function. If it is the same graph/curve for an individual or small group, then it is the same over T.

Further update: The velocity problem is exemplified by the graph you just providedl Given the variabiltiy, draw a line from the bottom and top points the range of values gets to be too great to continue it with any meaning. What I noticed is the line goes downwards. ????

[Peter McDonough]

DS and DT is simply a ratio of two different rates. All we need is a representation of how one's velocity changes over time and then one can figure everything else out.

I believe that finding this curve is impossible though.

The C and the 14 are unrelated. The 14 was just a part of the velocity equation Roman stipulated. The equation for others may involve a t cubed or rooted or a t cubed plus t rooted. The C is just a marker acknowleding that there are infinite solutions to a differential equation, and depends completly upon the initial value, and won't change for different people.

What changes is the velocity equation, and as such, their constant of integration depends only upon whatever inital value they plug in.

[b d]

I updated my previous communication, just saw yours. Agree, for now anyway, thus I can actually go to bed. Thank you for the dialogue and thought, most persevering and percipient.

[Roger Caffin]

I do hope no-one takes any of this seriously!

OK guys, now try to find the analytic solution to an elliptic integral.

[paul johnson]

Roman, see what you started!!!

BD, your humor was immediately apparent, either that or that lengthy appeal brief you were working on fried your gray matter. I'm still trying to figure out if CP knew from the outset that you were joking around. I think that he knew, but chose to deal with it as if it were serious, but i don't really know for sure.

I'd really like to see more from CP, who to my way of thinking is NOT very crazy at all. Good job, CP.

Oh, and PETE, you don't mind if i call you Pete, do you? (or is it Mr. Pete?), one word of advice, quit while you're ahead. BD, is a lawyer. You simply can't win with a lawyer. Even if you do, they just file another appeal! I really enjoy talking to attorney's, but they're a little like the Hydra that Heracles/Hercules fought. Slay one head and two heads pop out to take its place – that's the way it is with their arguments, slay one and they come back with two more – clever folk that they are.

[Peter McDonough]

Only if its definite Roger….

:D

[Peter McDonough]

Oh and one final quick reply—the velocity vs time graph was just an example that may or may not apply to anyone's walking speed.

The variability of the velocity is not a problem. Just think of that graph as one of you and your car, except without stoplights. All that graph is giving us is different velocities at different times. We use integration to stack an infinite number of rectangles under the line to calculate the area under the curve, thus giving us the total distance covered.

[b d]

pj, et al. –

Quite right that Crazy Pete is not crazy .. that is certain. In fact, I wish he could graph and create algorithims for law office scheduling efficiencies.

Ditto on the lawyers, hopefully trekking around will curtail the appeal upon appeal phenomenon, because of physical exhaustion if not new found common sense. Hope springs eternal.

I'm not even going to touch Rog's suggested new math project, leave that entirely for CP and RD. As Roman said earlier, what a really great group of amazing people hang out in the SUL/UL/L realms. Back to the trenches for me, may all have a great day and lighten their loads.

[Roman Dial]

So here is a graph of the model's solution plotted against the Arctic 1000 mileage data

[Roman Dial]

And here is the best fit which gives a slightly different relatiosnhip between walking speed and day as

ds/dt = 19.9 + 0.52 t.

Notice that around day 10 we drop below the "average" and again at day 22.

At day 6 and 7 we hit the really good ridge walking.

And to keep on thread: these data were recorded on 1:250,000 scale USGS paper maps with a pencil stub.

[b d]

Roman,

(Wrote this before the second graph, still looking at it.) Did you have a goal of so many miles a day? The points and curve are very 'symmetrical' or consistent.

At the end what caused the jump in miles per day, or it appears there was an increase over the norm? Did the terrain become easier to negotiate or did you travel longer hours?

IMO this shows the trekkers were in good shape and that the terrain didn't matter, they made their 20 or so miles a day, regardless of the time it took on a daily basis. Is that true or was the hiking time the same, you mentioned something about that? With you guys all being experienced and in good shape it may be that you covered the 20 or so miles regardless of terrain or elevation shifts.

A model of my distances per day would definitely show that I cannot get that far over rough terrain or an elevated hike in one day at a consistent rate. What I see this as showing is that trained, in shape trekkers can create a consistent day to day distance over time, regardless of terrain. A real goal and way to measure my success, also shows a strategy for covering the distance in a given time.

What you and CP are doing is very helpful to me in evaluating my own performance, health, fitness, and concept of what I am doing "out there."

[Roman Dial]

BD,

BTW you had me laughing out loud with your comments and the exchange with CP. It was as if you two had orchistrated a straight man funny man act. Good fun. I was L-ingOL (or is it LOL_ing?) over your convolutions.

IN the two graphs the dots are what we actually did and the curves are the models.

The only goal we had each day was 12 hours of travelling, although we did go more than 12 and less than 12, of course.

The points climb more steeply at the end because with a lighter pack we could go further, both faster and longer — that was UL backpacking at its best.

Of course the terrain did matter — swimmimg rivers and snowy passes slowed us down. So did bear detours and tussocks. But the big picture shows a consistent, but somewhat weak, trend of going farther, faster as the trip progressed, not just becasue we were stronger, smarter, more driven (because we also had less chocolate, more difficult travelling, and more sore feet) at the end, but because we could move more easily — hence farther each day with the same amount of overall effort.

If you were to go on a long hike and record the distance walked each day and plot the distance from the start, your curve would likely look similar. Your initial distance per day may be less, so your average distances per day would also be less, but the slightly upward curve to the data points would be there for you, too, I reckon.

Anyway, thanks for indulging me!

Some like to tinker with gear, I like to tinker with the relationship between my body and Earth's landscapes and to interpret them in a variety of ways — analytical as well as emotional.

[Author]

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