That seems a bit over the top. It seems obvious that a relationship between scope and depth is meaningful. How is that "winging it"?
I mean even you "just" using your chain to hold you..... do you put the same amount of chain (rode) out in every depth? Do you bother with an anchor on the end of it? (I think that even chain can become all taught in enough wind but it sounds like not yours?)
So you just put out X feet (or meters) of chain no matter the depth. Same every time?
The whole scope idea is based on wind and purpose.
So 3 : 1 is for a short lunch or swim break with no wind.
A 5 : 1 scope is for an overnite with no wind
A 7 : 1 scope is for an overnite with winds up to about 20 kts
A 10 : 1 scope is for an overnite with winds over 20 kts.
Now when you start calculating chain length in those situations you will find out that you can get close to what the math would also tell you.
As stated before the stronger the wind or the current the higher the counteracting force needs to be and that means more chain.
Let's say you want to anchor in 5 mtrs of depth and there will be max 2 Bft, then the wind force will be 6 N/m2. If you then have 25 m2 of frontal surface, the total force on your boat would be 150 N. My anchor weighs 50 kg, so that is already 500 N and the required chain to get to 5 mtrs of depth would be around 8 or 9 mtrs. At 4 kg or 40 N per mtr that would make it 320 - 360 N, which means that I would have a total of close to 850 N to counteract the 150 N the wind would put on my boat. So in theory I would be fine with just laying the anchor on the sea bed, not even setting it. And that is then for no waves, no current and no change in wind speed. As soon as there is a current or waves or a change in wind speed, you would need to change your calculations.
So for safety purposes all calculations start with a minimum of 4 bft.
If we would anchor at the same windspeed in e.g. 15 mtrs of depth then automatically we will have more chain, otherwise we don't reach the sea bed and we would just be floating. So that would be around 19 mtrs of chain for my boat, which is 76 kg, plus the 50 of the anchor, makes a total of 126 kg or 1260 N, which would be more than enough to counteract the 150 N the wind will put on the boat.
The problem is that in my area there is no accurate forecast, so I will never anchor with a guaranteed 2 Bft in mind. In 99 % of the time the forecast is not a little bit off, it is way off. There have even been times that 2 Bft was forecasted and in reality I had 12 Bft.
So the answer to your question is that in theory you could anchor based on the forecasted wind and if you would calculate it correctly you would be safe. However, any moment the forecast is off you could be dragging your anchor, since you have calculated for much less wind.
When you would calculate the required chain for 0 speed, 20 kts and 35 kts you will find out that you will be close to that 3 : 1, 5 : 1 and 7 : 1 . And that is also the explanation where that scope came from, since it did not get out of nowhere. The problem however is that it does remain a simplification of the math. May work 50 or even 90 % of the time, but it does not work 100 % of the time. Math does work 100 % of the time, that is the beauty of math in this case. It will tell you exactly what you need for a certain windspeed and current.
The main problem is that the forecasts are often way off and so it is up to you to choose wisely. Good part is that if you calculate for max 8 Bft and you find yourself in 10 bft you know instantly what to do. You don't need to raise the anchor, you just drop more chain and the dragging will stop.
As for taking the holding power of the anchor into my calculations ?
No I don't and I don't do it for a reason.
The anchor will have holding power as long as the shank is not raised more than 25 degrees off the sea bed.
Let's say the anchor has 2500 N holding power when it is dug in.
As soon as that shank is raised to 20 degrees that holding power is now only 1250 N.
If I would need e.g. 4000 N to prevent the anchor from dragging and the chain would provide 1500 N, i would be safe if the anchor shank is level with the sea bed. But as soon as that shank starts to raise I will be below that 4000 N and ultimately I will drag that anchor.
So in my calculations I will calculate the weight of the anchor, but not the multiplying factor which you get through anchoring in a certain type of sea bed. If the anchor has holding power then that is a bonus for me, it is an added safety, but not needed.
I have anchored in small coves where I dropped the anchor on one side of the cove, came 30 mtrs reverse to the other side of the cove and tied of the stern to the shore. Due to the depths of the water I had dropped the anchor in 8 mtrs, but in the middle of the cove it went down to 18 mtrs and then back up again to my boat. In total I did not have 30 mtrs out, but over 50 mtrs. My stern lines were as tight as could be and then I found out that the anchor was lying on some rocks, had not dug in at all. So all I was doing was lying on the weight of the chain and we stayed like that for almost 5 days. Most wind we had in that cave was perhaps 15 to 20 kts, but was no problem at all. I knew I had more than enough chain out to counteract the force that 15 to 20 kts could put on the boat...........even with the shore only a few meters away from the stern.
