1665

Item 1665

MAKE: ElectrotorPlus ~ MRGA - Motor - Mechanical Specifications

Much of the following has been coded into Access database; ELECTROTOR - Gyro and Gyro Data forms

Initial Proposed Constants:

Constants:

The weight of rotor is 10 lb [4.55 Kg] as a guess, at this point in time.

This suggests 2 x 10 = 20 lb for 15 HP motor. I.e. 1.5 lb per HP.
Gears etc will make up part of the mass. Planets are turning in wrong direction for gyroscopic precession.

Mass [M]= Weight = 10 lb = 4.54 Kg. Weight [lb] / 32.17 [ft/sec²] = 10 / 32.17 = 0.3108 slug.

About Z-axis:

Radii of the rotating mass of the rotor is OD = 13" = 0.3302 m and ID = 11.5" = 0.2921 m.

Radius of Gyration of the rotating mass of the rotor ~ From Machinery's Handbook, p. 143: [k₀]&[R_g] : [R_gZ]= sqrt (J_M / M) = sqrt (0.441 kg-m² / 4.55 kg) = sqrt (0.0969) = 0.3113 m. Interesting, re previous line. I would have thought that it would be further out than the mid-mass radius of the wheel.

Moment of inertia [J_MZ] [JmZ] for hollow cylinder. Reference to axis A-A ~ From Machinery's Handbook, p. 144: J_{M =}¹/₂M(R²+ r²). ~ R is the outside radius of the cylinder. r is the inside radius. J_{M =}¹/₂M(R²+ r²) = 1/2 * 4.54(0.3302² + 0.2921²) = 1/2 * 4.54(0.109 + 0.0853) = 0.441 kg-m² = 0.0903 lb-ft².

WAS Angular velocity: [ω_Z] [OmegaZ] (of motor's rotor) = 6,000 rpm * 2 * pi / 60 = 954.5 rad/min. = 628.6 rad/sec.

NEW Angular velocity: [ω_Z] [OmegaZ] (of motor's rotor) = 19,205 rpm * 2 * pi / 60 = 954.5 rad/min. = 2,012 rad/sec.

WAS Angular momentum: [L_Z] [LZ] = J_MZ * ω_Z (of rotor) = 0.441 kg-m² * 628.6 rad/sec = 277.2 (kgm²s^-1).

NEW Angular momentum: [L_Z] [LZ] = J_MZ * ω_Z (of rotor) = 0.441 kg-m² * 2,012 rad/sec = 887 (kgm²s^-1).

Variables:

About X-axis and about Y-axis:

Half length of motor's rotor; [l] = 0.127 m

Moment of inertia for hollow cylinder [J_MXY] [JmXY]. Reference to axis B-B ~ From Machinery's Handbook, p. 144: J_MX = M ( (l²/3) + ( (R² + r² )/4) ). ~ l is the half length of the cylinder. J_M = M * ( (l²/3) + ((R² + r²)/4) ) = 4.54 * ((0.127/ 2)² / 3) + ((0.3302² + 0.2921²) / 4)) = 4.54 * (0.00134 + 0.04859) = 0.2267 kg-m²= 0.0464 lb-ft².

Re [JmXy]; It appears that J_MX = M ( (l²/3) + ( (R² + r² )/4) ) can be replaced by J_MX = M ( (l²/3) + ( (R_gZ⁴ )/4) ) and get the same answers. I am therefore using the later algorithm, since [R_gZ]is a variable in the database form.

Ie. using '=[M]*((([RgXY]^2)/3)+(([RgZ]^4)/4))' instead of '=[M]*((([RgXY]^2)/3)+((0.3302^2*0.2921^2)/4))'.

Radius of Gyration of the rotating mass of the rotor ~ From Machinery's Handbook, p. 143: [k₀]&[R_g]: [R_gXY] = sqrt (J_M / M) = sqrt (0.2267 kg-m² / 4.55 kg) = sqrt (0.0499) = 0.2235 m

~ From here go to ElectrotorPlus_Gyro.html - Calculations ~

Place database forms Gyro and Gyro Data on Web page and link to it.

Flywheel:

For information about Flywheel and limitation see; Machinery's Handbook, page 171.

Cooling:

Consider pressurizing the gearbox or the area at the top of the motor's air gap sothat there will be an airflow downward through the air gap to the outside air..

Last Revised: August 6, 2008