Planning for current Chesapeake and Delaware Bays

[Pandor]

Originally Posted by sailingharry

Hmm... but a cursory review of the diagram (still on screen, I need to print it out and play with parallel rules) leads me to believe that the diagram is not as smart as it seems, and is less functional for slower boats (ie sailboats) and higher currents (ie, the DE Bay). It seems that the input is NOT SOW, but rather SOG. If you plot a line up the Chesapeake at 12 kts (their example) and run in a flood, you will be in a flood all the way -- but if you run in an ebb, you will be in an ebb the entire way and it will take the same steaming time. I need to print it and play!

No. The speed scales on the right hand side of the current diagrams use speed through the water. You are correct, that the Chesapeake diagram is not as useful as the Delaware posted below. So let's use this as an example. If one intends to transit the Delaware Bay southbound from Chesapeake City, that is a trip of 60 NM (10 NM from C. City to Reedy Point canal exit and then 50 NM down the bay to the Delaware Bay entrance).

Let's say you are in a sailboat that goes 6 knots in still water. By using parallel rulers to transfer the 6 knot line as evenly across the "EBB" portion of the diagram from Reedy Island to DB Entrance, you get an optimal departure from Reedy Island at 1h after Ebb begins at DB Entrance. Today that would correspond with 1212. You will also find that arrival at DB entrance will be at 3h after flood begins. Today that would be 2042. Therefore, total time to transit 50NM from Reedy Island to DB Entrance is 8h 30m which equates to a SOG of 5.9 knots. This is slower that the 6 knots if there was no current due to the adverse current you run into prior to the ebb and after.

But as a worst case scenario, you can transfer the 6 knot speed line as evenly across the Flood portion of the diagram as possible. You will find that if you do this the trip down the bay at 6 knots through the water will take 9h or an average SOG of 5.5.

Back to the optimal departure from Reedy Island at 1h after ebb at DB Entrance. If you back up 1h 40m to account for the 10 NM canal transit from Chesapeake City to Reedy Island, you would have had to leave C City today at 1032 to get to Reedy Island at the optimal time. High Tide at C City today was at 1136. So this method seems to differ by 1h from the old rule of thumb to depart Chesapeake City at high water.

Edit: I had one of my times wrong and went back and corrected.

[Pandor]

Originally Posted by Pandor

No. The speed scales on the right hand side of the current diagrams use speed through the water.

I stand corrected. After looking back at the diagrams, you are correct that the speed scales seem based on SOG. Here is the Delaware Bay diagram with four runs and times based on today's currents at Delaware Bay Entrance. Red and Blue are southbound and Green and Orange northbound. All runs take approximately 8.5 hours except the red one which takes closer to 9 hours. But I guess you could average the current values that are along the lines and then adjust your speed line inclination to account for the current. Then you'd have to redraw the new line on the diagram corresponding to your new SOG.