Aha! RMS velocity and the old 1.414 got me again!markrob wrote:The relationship between velocity, excursion, and frequency should be:
E = V/(2 x pi x f) or V = E x 2 x pi x f
Where V is the velocity and E is the excursion. Note that the 5 cm/sec reference is the rms value not the peak (rms is what your voltmeter reads as well). The peak 1.41 x RMS (Square root of 2). So for 5 cm/sec rms, you really have 7.07 cm/sec peak velocity. If you put that into the first formula, you get E = .001125 cm peak excursion. Peak to peak would be twice this value or +/-.00225 cm = +/-.0225mm..
Thanks for the equation Mark!
I’m going to need to do new calculations. The plan will be to update the diagram with correct values and equations shown so that future readers don’t go down the wrong path. Garbage in, garbage out…
As far as that new test jig goes:
As it is now, there are opposing drivers. They are 1 foot apart. The test fixture is non magnetic (aluminum). The connection between the drivers (right now) is also non magnetic Carbon Fiber. There is a support piece in the center that is also non-magnetic.
The idea here is that I can drive one side, and measure on either side knowing there is no significant magnetic coupling between the two ends. Also, driving one side will result in a mechanical load being applied on the other side. If the load offered by the spring only on the opposite side is not sufficient, the driver on the opposite side can be fed with an out of phase signal with adjustable intensity to increase the load on the opposite driver.
The push rod between the two drivers can easily be replaced by one that has a section of magnetic (like spring steel) coupled to a section of nonmagnetic material (like carbon fiber) to simulate only ½ of the 45/45 drive section in the head while still proving the ability to apply a mechanical load.
The test fixture allows mounting a measuring device with different orientations and distances to the driver or push rod using mounting adapters as needed. The fixture also allows shielding to be inserted, formed, oriented, and the effects measured.
Even though the last round of calculations were brutally flawed, it did clue me into another issue with this head design. In the diagram shown earlier, the driver connection noted as “D” implies the two 45/45 drivers are connecting at a single point (as they were in my other head design). On this head, the two drivers connect slightly offset at the torque tube. Not by much, but enough to explain why the sine wave cut is not symmetrical. Here is an earlier shot of the construction to refresh your memory... I'll measure it but I think it was 1.5mm so the two drive push rods would clear each other at the "X marks the spot".
And as Mark pointed out as something to watch out for is significant levels of odd harmonics showing up when doing a 1khz mono test. I remember seeing that in a 1K test that I may not have posted in hindsight... I also saw a non-symmetric looking waveform and didn’t make the connection other than that possibly my amp must have an imbalance (which it also does). All is not lost… I just need to drive one side hotter to make sure the cutting movement is balanced.
So here again is a photo of that 1khz groove from a post from Dec 19th of last year and none of us spotted the non-symmetric nature of that waveform! Shame on me for not making the connection to Mark's comments... But shame on you guys for not noticing either! Look at the groove. The first portion of the waveform is shorter than the second part. Minimum to maximum is shorter than maximum to minimum...
So I’m going to open this up to discussion until I have time to play with this….
What do you guys think?