My maths and physics are rusty but if you mean is the force on the anchor
the same as on the stem when the boat
is being held by a person ie the horizontal force due to wind
, the same as on the warp and therefore the anchor
Drawing a triangle of the depth
and the rode
the third line is the horizontal force equal to the wind
effect since the boat
is not moving.
The force on the warp is the hypotenuse and is clearly greater than the horizontal (the lengths of the lines being the horizontal and vertical forces relative to each other.)
We have two lengths usually the depth
and the length of scope
and need to find the length of the third side the horizontal . We can calculate the angle of the scope
to the bottom from the angle whose sine is depth/scope.
The resultant force on the warp is the horizontal force/cosine of that angle say a.
On my calculations the force on the rode
is 106% of the horizontal force at 3 times depth 103.3% at 4 down to 101.04 at 7 times depth.
So there is a difference but it is not great.
The vertical component that is the force lifting the anchor or conversely pulling the boat down is about 35% at rode/depth of 3 down to 14.5% at 7.
That is simplified by assuming an all rope
rode to avoid the complications of changing angles of a cantenary of chain and the weight of the chain. Of course when the chain is tight it has the same angle as a rope
. The forces on both the anchor and stem of the boat are both vertical and horizontal so will be greater than the horizontal force of simply holding the boat at the dock
by the bow.