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Runoff Report 1999..... The Time Span


As referred by previous sections the runoff time span is a very important indicator for the evaluation if flooding can be expected or not. The time span is mainly influenced by

   - the expected mean temperature during the runoff and
   - the mean height of the snowpack.

The height of the mean snowpack can be calculated using the snow pillow data from different altitudes. The links section of this web site shows the online access to such information for this region.

The expected mean temperature is a combination of long-term weather forecasting and calculated data from previous years.
Here some examples for previous years, based on my own recordings (all from March 1st to October 31st, Sicamous, lake area):

Year Mean Temperature in C
1999 14.5
1998 17.9
1997 15.1
1996 13.6
1995 14.7

All factors are in close relation to each other. Changing one factor will change the time span. For example, higher mean temperatures will shorten the time span, while a higher snowpack will extend it.

The total water supply for the runoff is a direct result from the actual snowpack. The potential water mass can be calculated into a water equivalent as done by Environment Canada for describing the snowpack. I will not do that because of too many unknown factors that influence those calculations, like snow density etc.

A long time span means that the fixed amount of snow water reaches the lake over a larger period, which allows more drainage and a slower increase of the lake level itself. The shorter the time span, the faster the total snow water potential gets into the lake. The lake raises faster, leading to an earlier and higher peak.

At the beginning of my research I tried to calculate that. But since I was not able to gain access to the necessary detailed data I had to abandon this attempt (as many others as well!). But for those who are interested in this subject I decided to show and explain my initial approach. Please keep in mind that the following formulas by no means reflect the reality. The formulas use linear calculations for the effect of the temperatures, which in my opinion is not correct. I just don't have the data to analyze this matter. Also, the results are not given in days, only in percent of the normal expectations. But it should give some ideas how the factors may change the time span. It's fun to play around with, therefore it's there.


Here is the first formula for calculating the time span:

Time Span (%) = Snowpack * 1999 Mean Temperature * 100

Normal Snowpack * Mean Temperature

Where:

  • Time Span
     100 % = normal
     100+ = shorter than normal
     1 - 100 = longer than normal)
  • Snowpack = height of actual mean snowpack
  • Normal Snowpack = height of normal mean snowpack
  • Mean Temperature = average temperature during runoff period
  • 1999 Mean Temperature = normal average temperature (1999 = 15º C)

 

Example:

  • Snowpack = 3m
  • Normal Snowpack = 2m
  • Mean Temperature = 18º C
  • 1999 Mean Temperature = 15º C
Time Span =

(3 * 15 *100) / (2 * 18)

= 125%

In this example the runoff time span would be 25% longer than normal.



To perform different calculations simply fill the boxes below and click on Calculate.
(Normal Snowpack and 1999 Mean Temperature remain!)
Snowpack (m)
Mean Temperature (C)


As described earlier changes in the time span effect the danger of flooding for the lake area.




The next formula integrates the time span to calculate the expected flood danger.

Flood Factor = Snowpack * Mean Temperature * 6

Normal Snowpack * 1999 Mean Temperature * (Critical Level - Lake Level + 1)

Where:

  • Flood Factor
     below 1 = no flooding expected
     above 1 = danger of flooding possible or expected
  • Critical Level = flood level, where spilling occurs, set at 349m ASL
  • Lake Level = initial or actual lake level (valid range: from 344m to 349m ASL)

 

Example:

  • Snowpack = 3m
  • Mean Temperature = 18º C
  • Normal Snowpack = 2m
  • 1999 Mean Temperature = 15º C
  • Critical Level = 349m
  • Lake Level = 346m
Flood Factor =

(3 * 18 * 6) / (2 * 15 * (349 - 346 + 1))

= 2.7

The flood factor of 1 would describe a normal year without flood danger. The result of the example shows a small chance of flood related problems. This chance will drastically increase when using a higher lake level in the formula.



To perform different calculations simply fill the boxes below and click on Calculate.
(Normal Snowpack and 1999 Mean Temperature remain!)
Snowpack (m)
Mean Temperature (C)
Lake Level (m)



Again, both formulas lack on precise data to be exact. They should only be used to get some general understanding of the mathematical approach in this regard. Maybe one day I will continue here, supposed the necessary data are made available more freely.


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