Torque transmitted through a helix angle?

[Hutch]

Hi all, my first post here,
I've succumbed to some help if possible, i know its wrong but it's getting my goat! now, and many times in years gone by.

I am building something as always, and reverse engineering exisiting drive components to select suitable new ones if that makes sense. In particular I am looking at a screw helix mathemtically (trying). or should i say just a screw.

what i want know is the relationship between :

the torque applied at a radius of a screw,
and the linear resultant force.

taking into consideration of course the helix angle, or i think, the helix angle. as that has a relationship with the force output, through an angle.

for instance, a screw thread with diameter 8 and a lead (pitch) of 2.5 between threads. using the trig formula

inv tan = opp / ADJ

and modifying slightly to add 'pi' , as the helix angle is using one full circumfrence of the screw against one pitch. and we are not working in 2d anymore.

inv tan = ( opp / pi.ADJ )

gives a helix angle of 5.68 degrees, great, the helix angle.

now,
what i would like to know is how does that helix angle now affect the output force as a multiplier, forgiving friction, with regard to what is input (torque/any) through the handle
.
i can see it is a ratio, and it changes uniformly, it is a decimal number, a mutliplier, how is the geometrical ratio now expressed?, how much force is output through a helix angle, any!.

 

[Hutch]

bieng as the screw moved 2.5mm of linear movement for a total circular movement of 25.1327412mm , is that not proportional to the output ratio in itself? 25.1327412/2.5 = 10.05 would then an input force be multiplied by this 10.05 ratio to give what's output?

i.e 10nm torque input on the short handle = (10N x 0.125M ) = 1.25Nm at the handle

multiplied by 10.05 = 12.56N output?

 

[ank_gl]

bieng as the screw moved 2.5mm of linear movement for a total circular movement of 25.1327412mm , is that not proportional to the output ratio in itself? 25.1327412/2.5 = 10.05 would then an input force be multiplied by this 10.05 ratio to give what's output?

i.e 10nm torque input on the short handle = (10N x 0.125M ) = 1.25Nm at the handle

multiplied by 10.05 = 12.56N output?

correct, if velocity ratio comes out to be 1/10 (use law of conservation of energy), the force at output increases ten folds

 

[Enthalpy]

I hope your spindle has balls. If not, friction consumes much more torque than pitch does.

 

[Hutch]

friction is omitted purposely from the assumption, (read 1st post) there are no balls at this stage, Its just for proving the maths, thanks by the way.

i find it interesting that force output through a screw can be calculated with the ratio of the pitch (distance traveled axially) to the circumfrence (work done in one turn), I guess those two factors produce a ratio, of course, and also a helix angle.

is there not a way that you can include the helix angle into a calculation? or is it irrelevant?