"How much HS download" is the start of a good question... but it's actually two questions, with two different answers.

Some people use "download" as shorthand for "negative incidence." In casual talk, that's OK. However, since loads are FORCES (measured in pounds, kilos, newtons, etc.) while incidence is an ANGLE (measured in degrees or radians), you can't answer the "how much" question without knowing if you are talking degrees or forces.

This, in turn, gets us into numbers. The numbers will be different for different airframes and rotors.

Say your rotor is five feet above CG. Say the gyro's CG is located on the mast and that prop thrust and fuselage drag are both right through the CG.

Say the rotor makes enough thrust just to hold the gyro up (one G) when the rotor flies at ten degrees AOA (after counting blowback) and 50 mph. Say the airframe is mounted to the mast at an angle that puts the frame perfectly level when the mast is angled back ten degrees to the horizon.

If the HS is mounted so that its AOA is zero when the frame is level, then the HS will generate neither a down-load nor an up-load force at 50 mph -- and that's fine, since the frame is level at that speed anyway.

Now say we speed up until the rotor can make one G at only 5 deg. Recall Chuck's cannonball-on-a-stick. The stick/mast (with the CG at its bottom) will try to adopt an angle of 5 deg. aft instead of ten. If there were no HS, it would succeed in adopting that 5-deg. angle.

At that angle, the frame will ride 5 deg nose-low. What if we don't like that? What if we want the frame level, like it was at 50 mph?

Well, the rotor needs to fly at 5 deg. Any more and the gyro will climb. Therefore, we want the frame now to fly level WITH THE ROTOR THRUSTLINE BEHIND THE CG. Specifically, the mast will ideally be at ten degrees while the rotor thrustline is at five. The rotor thrust will be pulling up on the tail.

Recall that the distance from the rotor down to the CG is five feet. A five-degree "delta" between rotor thrustline angle and mast angle means that, at the CG, the thrustline is about 5" aft of the CG. If the gyro weighs 1000 lb., the tail-up moment as long as the frame flies level is 5000 in.-lb. or 417 ft.-lb.

Therefore, to have the frame fly perfectly level at that higher speed, the HS must create a tail-down moment of 417 ft.-lb. For a 6-foot tail arm, that would mean a download force of 70 lb.

Here's what Vance may be getting at: IF you insist on ZERO change in frame angle with increased speed, a fixed HS can't do it. Our fixed HS will create ZERO download as long as the frame is level. The frame will have to ride a LITTLE nose-lower at high speed in order for the HS to "bite" and limit the nose drop with increased speed.

For example, if the HS is eight sq. ft. on a six-foot arm and our new higher speed is 100 mph, the HS will make roughly 2 lb./sq.ft. of download at -1.5 deg. AOA. 2 x 8 = 16. 16 x 6 = 96 ft.-lb. Meanwhile, the rotor thrust's aft displacement at 1.5 deg nose-down on the frame is .087 foot. At 1000 lbs. rotor thrust, that's 87 ft.-lb.

So, with that particular set of numbers, the gyro will ride a little less than 1.5 deg. MORE nose-down at 100 mph than at 50.

And, yes, you could get rid of that 1.5 with adjustable HS incidence.

If pod drag isn't centered on CG and/or increases exponentially with negative AOA, then you need a powerful HS. A flight-adjustable one would, again, limit the changes in deck angle, but the more important thing is that the pod not be able to over-power the HS, adjustable or not.

Trig buffs will note that I've simplified things here by treating rotor thrust as lift (thrust is actually slightly higher) and arc distances as chord distances. With the small angles we are dealing with, these approximations are pretty accurate, though.