Now a question for the boaters / marine experts: today I want to go to a spot where the ideal tide would be about a 2.6, whereas the low will be a 1.8. So if I recall the rule of twelfths correctly, to calculate when it will be this height, you divide the difference between the morning's high of 8.8 and the low of 1.8 I.e. 7/12 or 0.58. 1 hour before low will be 1.8 +0.58= 2.38 and 2hours 2.38+ 2x0.58= 3.54 so my ideal height would be about 1 1/4 hours before the low, I.e. 2.38 + 0.29= 2.67? There's probably a way of calculating it exactly using algebra but I just wanted to check I'm roughly right. And yes, I do realise that although the height may be ideal for weed beds etc, it won't be the same degree of slackness, if that makes sense, as if the low was actually a 2.6.
Or am I talking a load of bow locks?
Or am I talking a load of bow locks?