How To Calculate Centrifugal Force 

Hi VictoryFirst! IF the Safety Limit is with respect to the 'Centrifugal Tensile Force magnets will have to withstand', THEN the following ADDITIONAL information is required: 1.) Mass of one magnet in kg. 2.) Radius length of magnet to bearing-center (plane-of-rotation) in meters. Note 1: Centrifugal Force in Lbf is equal to: [(Lb Weight of one magnet / 2.2) x (radius feet / 3.28)] x [(RPM / 60) x (2 x 3.14159)]^2 / [4.45]. Note 2: The fastening method for retaining magnet must overcome Centrifugal Force times 2.0, to meet minimum Safety Limit. Note 3: The Young's Modulus (E) is equal to Tensile Force in Lb (F) divided into the Cross-Section-Area of Fastener in Sq In (A) which is equal to Stress, then divided into Elongation (dL) divided into the original length (L) or Strain. Stress / Strain = E. The (E) value will determine the tensile force limit prior to Yield and Ultimate Failure per Square Inch of Cross-Sectional Area. Note 4: Various materials, whether steel bolts, soft iron magnet interference-fit, aluminum, etc, have published their Young's Modulus limits, including stretch or bolt-stretch Yield Limit (F). Note 5: Find out the weight of one magnet, and measure its distance from the center of rotation (radius). You can submit the data in Lb Weight and Feet Distance if you would like. Converting these values to meters and kg can be done later during the calculation. Thanks VictoryFirst. Merry Christmas and a Happy New Year to you too!

Thank you for your response on my "Find the limit of safety" on my heat machine I am trying to make happen Mark. I will work on making this happen and touch base with you more if I need guidance Sir. Peace to you Sir. vf. P. S. The magnets will be held in place by just one bolt since the disk Neo magnet will be in a pocket and some epoxy to take up any clearance issues Sir. I do not have any high technical ways to verify what is safe or what is not. I just do not want to have an explosion and see the unit flying around till the electric cord gets yanked out of the wall socket. The motor is a 3 H.P. with a 3450 or 3600 R.P. M. 240 volts. Good day fella too.@@markscythian7179

@@victoryfirst2878 Drilling Soft Iron or Neodymium magnets is difficult and tedious. This may distort the Magnetic Lines-of-Flux into infinite feedback loops. Drilling magnets is not recommended. An acceptable method to fasten magnet(s) onto rotor could include top-down slide-insert and Sheer-Fit. A non-magnetic Hard Chrome top-plate could then be installed plush for magnet containment and held in place with a Cam-Lock. Simple yet effective. If sufficient rotor thickness is allowed, centrifugal force should not overcome sufficient thickness. Rotor mass, diameter, and RPM are logarithmically proportional to centrifugal force. IF you are building a 'Magneto-Heating-Element' to harness high voltage impulse DC, Armature Step-Up Coils/Inductors or Step-Up Transformer(s) may require Bonding-Straps to support electrical and thermal grounding. Electrical and Inductive grounding along with Heat Sink(s). Armature Coils are (practically) R-L-C DC Circuits. Initially, Rotor/Stator relationship is Sinusoidal and rectifies itself at the Armature Coils. Ceramic heating elements and ambient air reflow have solved many problems with respect to Heating Efficiency. A heating system commonly used in aerospace/defense applications, now slowly moving onto the civilian market.

Thanks Mark. @@markscythian7179

Centrifugal Lb Force (Cf) is equal to: [(radius in meters) x (mass in kg)] x [(6.28) x (RPM/60)]^2 / [(4.45)]. Centrifugal Force is exponentially greater than all other FORCES combined, that act upon a propeller during flight (Thrust Bending, Aerodynamic, Drag, Torque, and Procession). Airspeed has very little effect compared to Centrifigual Force. However, Airspeed has more to do with Propeller Drag Force, not Centrifugal Force.

Hi. Differential Calculus based Geophysics modeling disproves the claims of flat-earthers. However, people are entitled to their beliefs even if they are not scientific. After all, truth is stranger than fiction! Thanks for your comment.

Yes it will. Provided the RPM differential across all pullies is accounted for. The Length (L) is the distances from the pulleys planes-of-rotation (center of their bearings) to the outer tip of pulleys. Each pulley's mass (m) in kg TIMES each pulley's radius length (L) in meters TIMES the parenthesis quantity of each pulley's (Revolutions Per Second x 6.28)^Squared, then DIVIDED into 4.45 to calculate each pulley's Centrifugal Force in Lbf.

Similar, but not the same. Half the mass of the wheel, its Radius and RPM are used to compute Centrifugal Force of a wheel. The wheel's outward tensile force at a given RPM will equal its Tensile Centrifugal Force.

Centrifugal Force is the outward force vector acting on a body in rotation in relationship to its plane-of-rotation and is a function of the product of its meters radius, Kg mass times radians/sec quantity squared for Newtons Centrifugal Force. Of course there are 4.45 Newtons to 1 Lbs of Force as the conversion into US units.

Thank you for responding. When you swing a ball on a piece of string your hand exerts a centripetal force on the ball, the ball reacts and "EXERTS" a centrifugal force on your hand thus you have an action /reaction force pair that follows Newtons third law to a T. A rotating mass cannot be subject to both a centripetal force and a centrifugal force, which is what you seem to be saying, since these forces would cancel each other out.

My wording in my previous comment was WRONG, thanks for the correction Frank! Correction: Force Vector (magnitude & direction) exerted outward by a rotating mass in relationship to the plane-of-rotation, as a function of radius, radial mass and (radians/sec)^2. In the case of let's say propeller blades in rotation then Static Centrifugal Force exerted outward by the given rotating blade(s) mass in the form of Tensile Force exerted outwards upon the blade span structure by the rotating propeller blade(s) mass.

Thank you for you response; nothing causes more confusion than centrifugal force, you only have to look on the internet or in dictionaries to see that most of them have got it wrong.

A rotating mass EXERTING an outward force vector as opposed to "a force acting upon a rotating mass" which is NOT correct. The technicalities of the wording are often over-looked. Writing a statement compared to one's thoughts aren't always the same! Thanks again Frank for the correction! - Mark -

Hi Bob. The 'arm-mass' of rotating solid disc (grinding wheel) from plane-of-rotation (bearing) will equal: [(5.0 ounces) / (16.0 ounces/Lbs)] = 0.3125 Lbs. Then divided into 2.2 to calculate mass kg: [(0.3125 Lbs / 2.2.)] = 0.142 kg. The diameter of 5.0 inches divided into 12.0 inches: [(5.0 in / 12.0 in/ft)] = 0.417 feet. Then divided into 3.28 to calculate meters: [(0.417 ft / 3.28ft/m)] = 0.127 meters. Then divided into 2 to calculate radius: [(0.127 m) / (2)] = 0.0635 meters = 'arm-length'. Mass of Arm = 0.142 kg. Length of Arm = 0.0635 m. Cp = [(15,000 RPM / 60 seconds) x (6.28)]^2 x [(0.142 kg) x (0.0635 m)] = 22,226 Newtons "centrifugal" force. 4.45 Newtons = 1 Lbs. [(22,226 Newtons) / (4.45 Newtons/Lbs)] = 4,994.6 Lbs "centrifugal" force or approximately 5,000 Lbs of "centrifugal" force. The grinding wheel disc is subject to a 'Shatter RPM Limit' if "centrifugal" force exceeds a specified RPM Limit (usually stamped on grinding wheel). 5,000 Lbs of "centrifugal" force are exerted radially across grinding disc from bearing to edge, for the given grinding wheel specifications you have posted.

@@markscythian7179 WOW! Thank you! You are AMAZING!

@@Bob-yl9pm You're welcome!

Hi. 88,219.6 Lbs of centrifugal force for which 2,000 Lbs equals 1 U.S. Ton, would compute as 44.11 U.S. Tons of Tensile Force. As long as the tensile force was distributed at a theoretical vector equivalent to constant-RPM uniform angular centrifugal force which contacted the entire blade starting from its plane-of-rotation, then the theoretical answer is YES. However, replicating an identical tensile centrifugal force distribution by hanging an object on the blade would include error corrections involving contact area and effective vector (magnitude and direction) of where the weight is placed.

Thank you crowviking for liking the video lecture on 'How To Calculate Centrifugal Force'. As for calculating the centrifugal force at the equator of the Earth the radius at the equator of the Earth is obviously greater in comparison to the radial points throughout the rest of the Earth. Apparently there is a 5 mile thick "bulge" at the equator of the Earth due to the greater centrifugal force. Taking the diameter of the Earth + 5 miles divided into 2 would establish the radius of the Earth at the equator. Calculating the RPM of the Earth approximately ((1/24) / (60)] followed by researching the planar mass of the Earth across the equator radial point would establish the data required to know the approximate centrifugal force at the equator of the Earth. I will look up the applicable data and post an approximate calculation soon. Thanks again crowviking!

Calculating the approximate centrifugal force in Lbs at the Earth's equator would require making this calculation for a sphere. The Earth is not a perfect sphere but is an eccentrically shape spheroid. An approximate calculation based on the longest radial line (equator) using the entire mass of the Earth would suffice. Earth's total mass = 5.97 x 10^24 Kg Equatorial distance: 6,378 Km = 6,378,000 m 1 rotation in 24 hours = 0.0416 rotation per hour = 0.000693 rotation per minute = 0.00001155 rotation per second [(2 x 3.14159 x 0.00001155)^2 x (5.97 x 10^24 Kg) x (6,378,000 m)] / [4.45] = 4.5 x 10^22 Lbs Centrifugal Force Approximately 4.5 x 10^22 Lbs centrifugal force at the Earth's equator

Mark Scythian that seems a very unusual calculation. You have calculated the Cf felt by a point mass the mass of the earth, sitting in the equator of the earth. That isn't the Cf of the earth (which is a weird question that doesn't make much sense). The earth is spinning around its centre. You could find the Cf on any part of it (pick a 1m.m.m cube of dirt and calculate the Cf) but there isn't really any total,Cf of the entire earth on itself. It would equate to close to zero. In the above calculation you have calculated the Cf of one propellor arm, assuming a point mass at the distance if the propellor tip. The actual,Cf would be far less as a lot of the mass is closer than the propellor tip. If you want to calculate Cf felt on the earth by itself, you would need to either do a integral across the radius of the earth taking into account the mass and radius at each point, or find the CoM of that slice of earth. It's a pretty meaningless calculation. I think this person was actually asking what Cf a mass sitting on the equator. What Cf would say a human feel at the equator. For 80kg person it is about 2.8N

One other thing you got wrong. You said centripetal force is the opposite force to centrifugal. That isn't correct. Centrifugal force is an apparent force that only exists in the frame of reference of the object itself - this is an accelerating frame of reference. In this frame if reference there is no centripetal force. Centripetal,force exists in the inertial frame of reference. Say looking at the propellor from outside the spinning system. Here centripetal,force acts inwards. It is the force causing the propellor to go in a circle (it is the force causing the continual change in direction - and acceleration) There will be an outward force at the hub equal to this inward force. It happens to be the exact same size as centripetal and centrifugal, but that isn't actually centrifugal force though we often call it that. If the hub failed the blade would go in a tangent to the circle at the moment of separation, not straight outwards. Centrifugal force is the force felt on the object itself and only exists in the accelerating frame of reference. In the inertial frame there is no centrifugal force It's a thing people have a lot of difficulty understanding.

an available calculator say 0.003g ir 1/294 th of gravity. about 7 ounces for a 140 pound person. www.calctool.org/CALC/phys/newtonian/centrifugal you can chose units so for radius I used 3959 miles, and for angular speed I used 24 hours per revolution.

True. Thanks for the correction! Correction: TOTAL WEIGHT from Plane-of-Rotation or (11.5 Lb x 2), then divided into 2.2 to compute mass Kg, this value then multiplied times meters length of radius from Plane-of-Rotation. This product then multiplied times angular velocity or [(RPM / 60) x (2 x Pi)]^2, to compute Cf in Newtons force. Newtons force then divided into 4.45 Nf/Lbf to compute Lbf Centrifugal Force (Cf).

Calculating the radians/sec is the first step: [(RPM / 60) x (6.28)]^2 = Rads/Sec Multiplying the radians/sec time mass & length of arm is the second step: [(Rads/Sec) x (Arm Kg) x (Arm Meters)] Now the centrifugal force in Newtons is known. Divide the Newtons centrifugal Force into 4.45 to know the Lbs Centrifugal Force. Note: The arm length is taken from the center of propeller hub to the blade tip.

I'm getting different results in an online centrifugal force calculator, I'm multiplying Rad/sec by arm Kg and arm meters and getting vastly different results.