Hello,
I am looking at DC motors for a turntable drive system. I am going to experiment with a rim drive system, and am therefore worried about radial load. Manufacturer's specifications on a few motors I am considering are in the 5 Newton range. Does anyone know what kind of radial load a rim drive setup would exert on the motor shaft? What would be the determining factors at play? Platter weight?
I am looking at DC motors for a turntable drive system. I am going to experiment with a rim drive system, and am therefore worried about radial load. Manufacturer's specifications on a few motors I am considering are in the 5 Newton range. Does anyone know what kind of radial load a rim drive setup would exert on the motor shaft? What would be the determining factors at play? Platter weight?
That makes sense, at least intuitively. Now, how would this translate in a rim drive system? Drive wheel tension against the platter?
You cannot answer such a question. The Designer (you) can choose how much side - pressure the wheel exerts on the rim. There has to be some or there will not be any drive. Usually a spring system will push the axle of the wheel against the rim, that spring force is designed by you - and could be adjustable.
So this is a mechanical engineering problem - draw out the layout, add some forces and calculate. I have not done any mech eng for 40 years so would be the wrong person to work with.
So this is a mechanical engineering problem - draw out the layout, add some forces and calculate. I have not done any mech eng for 40 years so would be the wrong person to work with.
I understand. My initial description also lacked pertinent details regarding what I have in mind. There will be no spring mechanism involved. Rather, the drive wheel--housed in an enclosure separate from the platter--will be pressed against the side of the platter. Examples of this type can be found in the Teres Verus system, VPI rim drive, and others.
I have no mechanical engineering experience. Perhaps I was hoping someone could suggest a straightforward way to calculate or estimate radial force in such a system. I also accept the possibility of there being "no easy way".
I understand. My initial description also lacked pertinent details regarding what I have in mind. There will be no spring mechanism involved. Rather, the drive wheel--housed in an enclosure separate from the platter--will be pressed against the side of the platter. Examples of this type can be found in the Teres Verus system, VPI rim drive, and others.
I have no mechanical engineering experience. Perhaps I was hoping someone could suggest a straightforward way to calculate or estimate radial force in such a system. I also accept the possibility of there being "no easy way".
How much of a push does the turntable need?
Unless the bearings are bad, very little.
Imagine tying a fish-scale to a string around the rim, and run away at a speed which results in 33RPM. It may take a good pull to get it up to 33. But that is very short-term, and the motor side-force rating is about long-term wearout, it will stand much more for a second or two.
I'm going to guess that platter, disk, and needle altogether "could" need an ounce of force at the rim to maintain 33RPM.
If coefficient of friction was 1.0, you would need >1oz of force on the contact point. Rubber tends to be CoF near unity, but a highly polished rim might need more. So say >1oz. What is that in Newtons? Maybe 0.6?
Look at it the other way. The motor side-thrust rating must be adequate to use the motor. While a shaft-drive might be zero thrust, small motors are USUALLY loaded with gears, belts, chains, or friction wheels. All of these produce substantial side-thrust. If the maker does not allow for this, unhappy customers will rock his boat.
Unless the bearings are bad, very little.
Imagine tying a fish-scale to a string around the rim, and run away at a speed which results in 33RPM. It may take a good pull to get it up to 33. But that is very short-term, and the motor side-force rating is about long-term wearout, it will stand much more for a second or two.
I'm going to guess that platter, disk, and needle altogether "could" need an ounce of force at the rim to maintain 33RPM.
If coefficient of friction was 1.0, you would need >1oz of force on the contact point. Rubber tends to be CoF near unity, but a highly polished rim might need more. So say >1oz. What is that in Newtons? Maybe 0.6?
Look at it the other way. The motor side-thrust rating must be adequate to use the motor. While a shaft-drive might be zero thrust, small motors are USUALLY loaded with gears, belts, chains, or friction wheels. All of these produce substantial side-thrust. If the maker does not allow for this, unhappy customers will rock his boat.
Last edited:
I am not sure I understand your question completely. Some rim drive systems use gravity by leaning the shaft/pulley against the platter. Teres Verus and Transfi are examples of this. Systems by VPI and TT Weights seem to couple the pulley without leaning it, leaving the motor shaft parallel to the platter bearing.
I have to experiment with this yet, but I imagine it requires a force similar to that needed to place proper tension on a belt.
It may be counter intuitive but contact area is not necessarily part of the equation. IIRC normal force multiplied by the coefficient of friction gives the driving force. It has been many years since statics so I might be mis-remembering....
If you arrange the motor on a swinging arm in the right orientation, the driving force will generate the contact pressure and the greater the resistance to turning, the greater the contact pressure. Arrange the arm in a different orientation and the reverse would happen so a jammed turntable would cause the drive to slip.
Member
Joined 2019
I would make the rim force adjustable (spring), start with some low value and increase it while the wow value becomes stable. For the measurements I recommend the smartphone RPM WOW Meter of Andrea Martignano. Measure WITH the stylus down on some high groove modulation levels LP.
The Martignano software: YouTube
The Martignano software: YouTube
- Status
- This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.