This is a topic with which I measured myself several times and in many ways because, as you can imagine, it is very important to leave on a trip provided with the appropriate technical equipment: backpack, boots, shoes, clothing, food, compass, maps, etc.. but also with an idea as precise as possible, about the time it takes. Unfortunately, the daylight hours are often limited, especially in winter, our forces are not inexhaustible and often return to our home is looked forward to our relatives. So, it is important to have a criterion for estimating at home the time our tour takes.
Really, there are several criteria and the first in which I came across was based on the simple difference in altitude assuming to climb up to 300-400 m / hour and go down to 500 m / hour; this is useful but it does not consider the path length. You understand that the same drop of 1000m results differently if we walk for 2 km (gradient 50%) or if we have to walk for 10 km (10% slope) for which it is useful to associate to the simple drop also a parameter that takes into account the distance traveled.
Some time ago, inquiring about this, I found a useful feedback in a couple of Web pages published by friends UpTheHill and giorgio59m that gave this interesting information: to bind together drop and route length they reached the concept of linear equivalent effort (SLE) adding up the kilometers to go and further kilometers obtained by dividing the difference in altitude by 100.
For example, on a trip, which we know the length of 12 km and drop of 800 m we can determine the linear equivalent effort SLE by using the simple rules:
SLE = 12 km + (800/100) km = 12 km + 8 km = 20 km
Needless to say, we speak of an empirical formula that relates a real mountain tour to a hypothetical trip that takes place on a plane road but of different length (greater).
Already obtained this value (SLE), each of us should know and apply their traveling speed to the trail so calculated to determine the time needed to complete it. In this practical case, if we know we can walk the SLE at a speed of 4 km / hour in the example above we actually will spend:
Time = 20 km / 4 km/h = 5 hours
At this point I asked to me a further question: "How to calculate the forward and back part as well as you often have to climb down?". I replied, imagining to use two SLE : the first for the part up (SLEup) and the second to be used for downhill (SLEdw). Now let's take a more concrete example. Imagine that you go up 800 m in 12 km and then descend from another way for 14 km.
Then we have a linear equivalent effort up:
SLEup = 12 km + (800/100) km = 12 km + 8 km = 20 km
and a linear equivalent effort for the descent:
SLEdw = 14 km + (800/100) km = 14 km + 8 km = 22 km
Now, to know the total time to be spent we have to apply two different speeds of travel, one for the climb (we imagine is 4 km / hour) and one for the descent (suppose to be 6 km / hour) and the game is made:
Time = SLEup / Velup + SLEdw / Veldw (a)
Time = (20 km / 4 km/hour) + (22 km / 6 km/hour) = 5h + 3h40' = 8h40'
The third question I have asked to me is: "What if there are any ups and downs involved?". The answer is simple: reduce our path to a series of segments up or down and use the formula (a) as a sum of as many terms as are the segments we considered applying for each term the speed of climb up or down as appropriate.
The formula (a) become more generalized as follows:
Time = SLEup1 / Velup + SLEup2 / Velup + …..+ SLEupN / Velup + SLEdw1 / Veldw + SLEdw2 / Veldw + …. + SLEdwN / Veldw
It sounds complicated but it is not; let's do the usual example.
We have to climb 1000 m in 12 km, 300 m down in 2 km, climb 200 m in 1 km and then 900 m down in 11 km. Then the segments are:
SLEup1 = 12 + (1000/100) = 22 km
SLEdw1 = 2 + (300/100) = 5 km
SLEup2 = 1 + (200/100) = 3 km
SLEdw2 = 11 + (900/100) = 20 km
Now we will apply the up or down speed for each segment depending on whether it is going up or down and therefore:
Time = SLEup1/Velup + SLEdw1/Veldw + SLEup2/Velup + SLEdw2/Veldw
which in numbers means:
Time = ( 22 / 4 ) + ( 5 / 6) + ( 3 / 4 ) + (20 / 6) = 10,41 hours
who converted to sexagesimal (just divide the centesimal minutes by 1.6) are:
Time = 10h25' (I must say….a tiring hike!)
What about now? What is the accuracy of this method?
Please note that:
I tried this method only on a hiking on "summer"; the use of snowshoes / ski / crampons can greatly changing the rate of ascent and descent. The method is applicable only in the hiking; vie ferrate and climbing in the strict sense should be considered with other parameters (eg, for climbing in two person out of IV / V rank I consider 15'-20 ' each rope length) and then go beyond what has been said here.
The methodology used is entirely empirical and somewhat individual, the speed of ascent and descent must be determined by personal experience.
In my case I checked in a number of trips that my climb speed is 5 km per hour and my down speed is 7 km / hour. I recommend: keep in mind that these speeds are "virtual" and they have value just applied to the "virtual" distances that are the SLE (linear equivalent effort).
The predictions that I do for a new hike, applying the above, are slightly large than the time actually used and include the time to take the pictures, to have my short stops, eating small snacks, excluding only the time for the main meal (one ½ hour).
The method is much better as good is the approximation with which we are able to determine altitude and length of each segment. They range from measurements made directly on the map with the aid of the wheel (not reliable) to data gathered by tools like Google Earth, Swiss Map, BaseCamp, Map Source or other cartographic good software (good reliability ).
When we speak of the "up" or "down" segment we mean the prevalence of ascent or descent; there will always be a bit of approximation but this allows us to reduce a trail to 2, 4 or up to 6 segments (very rare ).
In any case, to make life easier, you can arrange with an Excel spreadsheet in which to place the above formulas. For those interested, I will be able to provide the model for the Excel spreadsheet that I use regularly.