@Stringway Official - thank you for a corrected drawing. Let's use that picture as a basis for analysis. This is not for the faint of heart, I fully realize they may not be too many readers left by the end of this post

Here are two drawings illustrating the forces in question, figure on the left when the arm is level, and figure on the right when the arm is at an angle. the formulas work for any angle, for the purpose of having practical numbers I've assumed the angle is 30 degrees. The drawing is to scale.

It should be obvious that your formulas given here

https://tt.tennis-warehouse.com/index.php?threads/stringway-–-information-–-questions-–-answers.673819/post-14435668 are incorrect. What you actually want as a proof of " the tension is the same for every angle" is the following.

From the torque equality equation:

W * a = F * b where

W - dropdown weight in direction perpendicular to arm a

a - distance from pivot point p to the center of weight (we are assuming that the arm itself is weightless). This never changes no matter what the angle is

F - Force applied perpendicular to b

b - distance from pivot point p to where force F is applied. This never changes no matter what the angle is

when arm a is level W is the actual dropdown weight, and F is actual Tension force acting on a string. As such we have:

W * a = Ts * b => Ts = W * a / b

when arm a is at angle A we have:

W' * a = F * b where W' = W * cos(A)

however in this position F is no longer equal to Ts. Instead

Ts = F / cos(A)

as such we get

W * cos(A) * a = Ts * cos(A) * b => Ts = W * a / b

using sample numbers given the values from the drawings we have:

W = 10, a=100, b=20.6 when a is level we have

10 * 100 = Ts * 20.6 =>

**Ts = 48.54**
when a is at 30 degree angle we have:

W' = W * cos(30) = 10 * 0.866 = 8.66

8.66 * 100 = 20.6 * F => F = 42.04

**Ts** = F / cos(30) = 42.04 / 0.866 =

**48.54**
So indeed the force acting on a string is _the same at every angle_. Right? Right?

Well, almost. As you can see from the figure on the right that is true

___only as long as the force Ts acting on the string is in the direction level with the ground_.__
But unfortunately that is _never_ the case in practice. On the actual Stringway machine the point where the racket's grommet ends up being is fixed. As such the direction that the force Ts acting (which is along the direction CG where the string is clamped (point C) to the grommet (point G)) changes as the arm position changes. That is unlike in a 'normal' dropdown machine where that Ts force is always along the same direction.

** In addition, with Concorde lift contraption that direction changes significantly.**
Judging from various pictures it seems that with Concorde lift the direction the force Ts is acting is about 9.5 degrees vs the level. With that, even when arm a is level, that direction is _not_ level. Based on the drawing on the left we have:

Tsc = Ts / cos(9.5) = 49.23 where Tsc is the tension acting on a string when the racket is lifted

Admittedly that is not that significant of a difference between 48.54 and 49.23, but still.

However the machine is very 'sensitive' to the change of angle of arm a. When the arm a is at 30 degrees vs the level, and the frame is lifted by Concorde, we have:

with Concorde frame is raised. The distance L is:

L=d*sin(9.5)=13.23 where d is the distance from the pivot point to where the grommet is along the line level to the ground. Given the scaling of the drawing d is about 80.17.

from basic trigonometry we have:

b=b'+b"

d=d'+d"

d"=b*sin(A) = 20.6 * 0.5 = 10.3 => d' = d - d" = 80.17 - 10.3 = 69.87

b'= b * cos(A) = 20.6 * 0.866 = 17.84 => b" = b - b' = 20.6 - 17.84 = 2.76

tan(B) = b" / d' = 2.76 / 69.87 = 0.0395 => B = 2.262

Ts' = F / cos(A+B) = 42.04 / cos(32.262) = 42.04 / 0.846 = 49.69 this is the tension force acting on the string when arm a is at 30 degree angle and racket

**is NOT** lifted by Concorde.

this is (49.69 - 48.54) / 48.54 = 2.4% difference, which is not too bad still

with Concorde frame is raised by angle C where

tan(C) = (L+b")/d' = (13.23+2.76) / 69.87 = 0.2289 => C= 12.89

**Tsc'** = F / cos(A+C) = 42.04 / cos (30 + 12.89) = 42.04 / cos(39.5) = 42.04 / 0.73 =

** 57.59** this is the tension force acting on the string when arm a is at 30 degree angle and racket

**is **lifted by Concorde.

(57.59 - 48.54) / 48.54 = 18.6% difference which is rather significant. It is actually so significant that it makes we wonder if that entire analysis is correct. In practice it means that when you 'think' you are pulling 48.54 of tension you are actually pulling 57.59 of tension. Granted, only some of the strings are pulled while the racket is lifted by Concorde. One can also notice that when racket is lifted by Concorde there's additional friction against the grommet (due to the string being pulled at an angle) so the fact that the machine is pulling at a greater force in that state is a 'good thing'. However estimating the actual 'contribution' of that friction is entirely different story altogether. At the very least it would vary depending on the strings properties anyway, so the 'true' tension that would be acting on a string within the frame (i.e. between the grommets) would vary.