Physics of a sprint start I am looking for an explanation to this illustration of a sprinter coming out of the blocks:

Apparently, the production of horizontal force produces a great deal of body rotation (R1) which would quickly rotate the body to an upright position if unopposed. So to counteract R1 the sprinter has to produce vertical force to produce a counter-rotation (R2)
I can see how R1 is a result of the torque produced at the hip joint during hip extension. But I really can't seem to grasp how a production of vertical force can produce a counter rotation to that.
Anyone who can help?
 A: Thank you for elaborating, Sammy. I think I am starting to grasp it. 
The resultant force vector, however, can't be equal to the horizontal force vector since the sprinter needs to apply enough vertical force not only to support his weight but also to raise his center of mass. 
So I am guessing that the resultant vertical force is RAISING the sprinters center of mass (while also producing a clock wise rotation) whereas the horizontal force vector HAS to produce a LARGER counter clock wise rotation in order for the sprinter to actually be able to reach the ground with his legs.
I also assume this is why the resultant force vector HAS to be BELOW the COM during acceleration. Because if not, in theory, the sprinter would only experience a translational displacement of his COM with zero net torque.
Does that sounds right?
A: "Rotation" in physics is an effect.  The "body rotation" which you refer to is properly called a "moment" or "torque".  "Moment" (= force x perpendicular distance from the pivot point) is the tendency of a force to cause rotation.  
The issue here is whether the body will topple clockwise.  To discuss this, we consider rotation about the contact point P with the ground.  
The weight W=mg of the sprinter, acting through the centre of mass CM, causes the clockwise moment about P, which you have labelled $R_1$.  The push on the sprinter from the ground acts through P so it does not cause any moment about this point.  So where does the opposing anti-clockwise moment $R_2$ come from?
The sprinter is accelerating forward.  This produces a fictitious backward force F=ma acting through the sprinter's CM equal to ma, his mass times his acceleration.  This is similar to the fictitious centrifugal force experienced during circular motion (which is an acceleration towards the centre of the circle).
$R_2$ is the moment of this fictitious force F about the contact point P.  As long as the sprinter is accelerating forward this fictitious force provides the anti-clockwise moment $R_2$ which opposes the clockwise moment $R_1$.  As the sprinter reaches cruising speed his acceleration 'a', hence fictitious force F, hence the moment $R_2$ all decrease.  To avoid toppling over the sprinter has to straighten up, which reduces the moment $R_1$ about P caused by his weight W.

The explanation you found on Reddit is also a good one.
The sprinter does not accelerate vertically so the net vertical force is zero.  Therefore the upward vertical contact force with the ground $F_y$ equals the downward weight $W=mg$ of the sprinter.
The horizontal friction force $F_x$ on the sprinter causes his acceleration $a$ where $F_x = ma$.  Since the vertical forces are balanced, the horizontal force is the resultant of all the forces.  The resultant does not pass through the CM of the sprinter.
Taking moments about his CM, in order for the sprinter not to rotate and fall over the clockwise moment caused by $F_y$ must be balanced by the anti-clockwise moment causesd by $F_x$.  Weight W acts through the CM so it does not cause any turning effect about the CM.
As the sprinter reaches his top speed, his acceleration $a$ decreases, therefore $F_x = ma$ also decreases.  However, $F_y$ must remain equal to $W=mg$.  To avoid toppling over the sprinter must straighten up.  This brings the CM closer to the vertical line through P, reducing the moment of $F_y$.  At the same time it increases the distance of the CM from the horizontal line through P.  The clockwise and anti-clockwise moments remain equal so the sprinter does not rotate and fall over.
