Phil Allen Posted November 3, 2014 Share Posted November 3, 2014 Ok folks,I'm not the mathematician that John Huth or Leon are, but I did spend some time playing with current estimations this stormy weekend. There are two basic "rules' for estimating a current speed from knowing the max current, time of max current and time of slack; the rule of thirds and the 50/90 rule. As I remember them, the rule of thirds gives the hourly average current speed through out the cycle between slack, max current and back to slack. Specifically, 1/3 of the max speed in the first hour, 2/3'in the second, 3/3 in the 3rd and 4, 2/3 in 5 and 1/3 in the sixth. The 0, 50%, 90%, 100% (50/90 rule) describes the actual speed of the current at slack, 1 hour into the flow, 2 hours into the flow and at 3 hours into the flow. All this is assuming a "normal" six hour sine wave shaped curve.So why this post? I made the silly mistake of not taking things for face value and actually calculating the average speed of a 1 knot current (my example) over the cycle at various time intervals. If I do the average speed from 0-1, 1-2, 2-3... (the rule of thirds case), I actually get 25, 70 and 94% of max current. In contrast if I average from 30 min before to 30 min after, I basically get the 50/90 rule (0%, 50%, 90% and 97% for the average speed at -0.5-0.5, .5-1.5, 1.5-2.5, and 2.5-3.5 hours.While for the rule of thirds, 94% and 70% are really close to 3/3 and 2/3, 25% is pretty different than 1/3. I used max current speed as 1 NM/hour for the math, but some real life situations currents could be 6 NM/hour or greater. So getting it right can be kind of important.So my question is, which 'rule' do folks use, how do they use it and have I made any major mistakes or bad assumptions? bestPhil Quote Link to comment Share on other sites More sharing options...
rfolster Posted November 3, 2014 Share Posted November 3, 2014 I might expect the numbers to be relatively close, since you are using two measurements to measure the same entity - current. However, the Rule Of Thirds is used for such things as making crossings where you need an idea of what the current will be doing over a period of time. The 50/90 Rule is used for a specific point in time, such as when going around a headland or some other pinch-point. The rules are used for different purposes, and I don't think that they are interchangeable because in most data sets, rarely will an average coincide with any one data point along the way. Quote Link to comment Share on other sites More sharing options...
Phil Allen Posted November 3, 2014 Author Share Posted November 3, 2014 Hey Rob-That was how I'd learned to use them as well, But to my question, the Rule of Thirds seems to do a worse job of estimating average current speed between 0 and 1 hour (for example), than 50/90 applied to the interval +/- 30 minutes of the hour. Quote Link to comment Share on other sites More sharing options...
JohnHuth Posted November 3, 2014 Share Posted November 3, 2014 I always called it the 'rule of 12ths', but I know what you're talking about. The way I look at it, the 1:2:3:3:2:1 sequence is just describing the rough shape of the sine curve by breaking it up into hour-long pieces - so depending on where you are within an hour (beginning or end), you probably need to tweak things. For example in the 0-1 hour interval, at the start, you have zero (or close to it) current, but at the end of the hour, you would have 1/3 of the max flow. In any case, in my last post, I tried to compare the rule or 12ths for tide height and compared that to what showed up as the actual prediction and got to within about half a foot for a 12 foot tidal range for each of my predictions - it worked out reasonably well, but these were for the actual end of hours (e.g. end of hour 0-1). To get anything better than that, I'd do an eye-ball interpolation and call it good. So, I guess we're testing out the efficacy of these things. I'm having more trouble estimating max current in some random place where there's not a NOAA prediction. I tried to do this for the entrance of Machias Bay, and I *think* I did OK - but I had to play around a lot with trying to estimate how much the current picked up between, say, Casco Passage, Moosabec Reach, and then Eastport - and take an educated guess. Quote Link to comment Share on other sites More sharing options...
rfolster Posted November 3, 2014 Share Posted November 3, 2014 Clarification:Rule of Thirds - Average current over each hour from slack (to calculate drift)Rule of Twelfths - Calculate tide heightRule of 50/90 - Calculate current speed at the end of each hour from slackPhil, it might be that you are trying to use the Rule of Thirds and/or the 50/90 across time periods that they were not intended for. Each rule starts it's calculation from slack, and calculates over each hour (Rule of Thirds), or at the end of each hour (50/90 Rule), but not in between. I think that trying to shift the time period over each hour marker will cause the "Rule" to fall apart. Quote Link to comment Share on other sites More sharing options...
JohnHuth Posted November 3, 2014 Share Posted November 3, 2014 (edited) I think you can use the 1:2:3:3:2:1 rule for current as well, you just have to change which hour you're talking about. Max current is just midway through the high/low cycle - they're both sinusoids and all that 1:2:3:3:2:1 does is give you an approximation of the sine (or cosine). Just offset one by three hours to get the other. But I don't want to confuse, I'm just responding to Phil's question "what do you use" - having one rule for me is easier and knowing how to apply it is what I do. Edited November 3, 2014 by JohnHuth Quote Link to comment Share on other sites More sharing options...
Phil Allen Posted November 3, 2014 Author Share Posted November 3, 2014 Ok, let's see if I can include the plot that raised my questions. What I did was plot out the a sine wave from 0-180 degrees (which is always positive) and just converted the axis from degrees to time assuming a 6 hour period from slack to slack. That's the black line in the plot. I also did a running average of the sine data in two different ways "the rule of thirds" way, where each point is the average of the 60 minutes of current (sine) data before it. That's the blue curve. I also averaged the sine data over a range from -30 minutes from the point to plus 30 minutes from the point, which is the red curve.As you can see, the +/- 30 minute average provides a much more similar to the instantaneous current curve (black) even though it represents an average of an hour of time. If you were to look at 1 hour, 2 hour and 3 hour time points in detail, it's exceptionally close to the 50, 90 100 "rule".So do I have a novel insight, or have I messed something up in my math?bestPhil Quote Link to comment Share on other sites More sharing options...
JohnHuth Posted November 3, 2014 Share Posted November 3, 2014 I'm not 100% sure that I followed what you did in the blue curve, but I think that what you did with the red curve should give you a better approximation because you're evaluating the change on either side of the hour, as opposed to taking it on a particular hour, which will give you a systematic offset in time, which seems to be the case. Quote Link to comment Share on other sites More sharing options...
Phil Allen Posted November 3, 2014 Author Share Posted November 3, 2014 Blue curve is the average of the black curve from 0-1 hour for point 1 (at 1 hour), 2 min to 1h2min for the next point, etc. bestPhil Quote Link to comment Share on other sites More sharing options...
rfolster Posted November 3, 2014 Share Posted November 3, 2014 Is it possible that tidal currents "represent" a sine curve, but they don't really perform that way in the real world. If you look at your zero hour, your plot is above zero for current. While technically the plot at zero hour is zero current, there is a current value change in the miniscule time that equals the width of your y-axis. Your sine plot does not take into account that slack water is not instantaneous - it is not a linear changeover from negative flow to positive flow as your mathematically plotted curve would suggest. Maybe this lag time is figured in by the Rule of Thirds, and why the modified 50/90 Rule does not work. Quote Link to comment Share on other sites More sharing options...
Phil Allen Posted November 4, 2014 Author Share Posted November 4, 2014 Hey Rob-For analysis sake, I modeled a perfect sine curve spread out over six hours. As I understand the "rules", the perfect curve doesn't have to be 6 hours. If that's the case (say a 4 hour slack to slack period), you'd just break the time into sixths. If the change in current speed has two peaks or isn't well modeled by a sine wave, everybody's projections fall apart.And I'm not sure why you say the modified 50/90 rule doesn't work? The traditional 0/50/90/100 rule predicting current flow at times 0, 1/3, 2/3 and 3/3 between slack and max current is derived from taking the sine of 0, 30, 60 and 90 degrees, and 0%, 50%, 87.5% and 100 % are the exact values. The modified time averaged (+/- 30 minutes) 50/90 method, gives average flows of 0, 50.2%, 86.9%, and 98.9% of max current. In contrast the average current flow calculated using the rule of third time intervals 0-1, 1-2 and 2-3 hours are actually 26.4%, 70.5% and 95.7% of max current.bestPhil Quote Link to comment Share on other sites More sharing options...
JohnHuth Posted November 4, 2014 Share Posted November 4, 2014 Quoth Phil"For analysis sake, I modeled a perfect sine curve spread out over six hours. As I understand the "rules", the perfect curve doesn't have to be 6 hours. If that's the case (say a 4 hour slack to slack period), you'd just break the time into sixths. If the change in current speed has two peaks or isn't well modeled by a sine wave, everybody's projections fall apart."And the rest. Yes, it's mainly a question of how you can approximate a sine curve - many ways of doing that - for me the 1:2:3:3:2:1 is easiest to remember, and I just have to know where to put the 1's and the 3's for tide height versus current, and then even interpolate. There are some interesting currents that are off the 6 hour (12minute) timing. Slack to ebb at Hull Gut is about 2 hours or even less. Quote Link to comment Share on other sites More sharing options...
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