This partial calculator converts selected mechanical horsepower and rotational speed into SI power, angular speed and ideal average shaft torque. It does not model a stake puller, hydraulic pump, cam, jaw, post or soil.
Stake Puller - Selected Shaft Power Ledger
Explore the ideal relationship among selected mechanical horsepower, rotational speed and average shaft torque.
Equation Used
- Horsepower and rpm are selected inputs, not measured engine-curve data.
- Horsepower means mechanical horsepower at 550 ft·lbf/s.
- Torque is the ideal average value implied by P=Tomega.
- No losses or transient behavior are included.
- No stake-puller mechanism, force, capacity or safety claim is made.
The secondary figure is an abstract shaft ledger; arrow length and rotation are illustrative only.
Selected shaft power ledger
Mechanical horsepower is converted to watts, rpm is converted to radians per second, and ideal average torque is obtained from P=Tomega. The calculation is an input relationship, not an engine performance curve or a machine rating.
What the calculator determines
Use it to inspect the ideal relationship among a selected power, speed and average torque. It cannot determine pull force, cycle time, capacity or suitability for a stake-pulling application.
Exact selected relationships
P=745.6999 HP; omega=2pi n/60; T=P/omega. The displayed pound-foot value is a unit conversion from the SI torque.
Selected-input example
At 4.8 hp and 3,600 rpm, selected power is about 3.579 kW, angular speed is 377.0 rad/s and ideal average torque is about 9.49 N m or 7.00 lb ft.
Not a stake-puller performance model
Engine curves, duty, clutch and belt losses, pump displacement, pressure, efficiency, cylinder area, cam geometry, jaw grip, post shape, embedment, soil resistance, side load, structure and safety are excluded.
Questions
Is this torque available at every operating point?
No. It is the ideal average torque implied by the two selected inputs; actual engine torque requires a manufacturer performance curve and operating conditions.
Does it predict pulling force?
No. Converting shaft torque into pull force requires a defined transmission or hydraulic system and losses.
Why is the device animation gone?
The source did not establish the depicted engine, pump, cylinder and jaw as one validated mechanism, so the candidate shows only the supported shaft relationship.
What are the calculation references?
NIST Guide to the SI, Appendix B.8 supplies the mechanical-horsepower conversion, and the U.S. Department of Energy UNH Turbine Design Report states P=Tomega.
Where actuator force enters the Stake Puller Mechanism design
When stake puller mechanism is driven by a linear actuator, the actuator force changes through the travel as the geometry changes. A comfortable static force number at one position can still leave the actuator overloaded near the start or end of stroke.
With leverage, force is traded against travel. A bracket change that improves force can also increase required stroke, slow the output motion, or move the actuator into a poor mounting angle.
A senior review would normally separate the required output force from the actuator force. Output force belongs to the mechanism; actuator force belongs to the installed geometry. Confusing those two numbers is one of the most common causes of undersized linear actuators.
- Calculate actuator force at the worst bracket angle, not just at mid-stroke.
- Check stroke, retracted length, and extended length against the mechanism envelope.
- Confirm the structure can accept the higher loads created by a stronger actuator.
For a FIRGELLI actuator application, run the FIRGELLI actuator force calculator, then compare available stroke, force, and mounting envelope in the FIRGELLI linear actuator range. If the goal is to replace an existing unit, use the linear actuator replacement finder before assuming the new actuator is a direct fit.
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