RedcrabX – Physics Functions

Reference for the Physics panel  |  Force and Work group

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Table of Contents

• Overview
• Centrifugal Force Solver – centforce
• Pressure-Force-Area Solver – pfa
• Work-Force-Distance Solver – work
• Power Solver – power
• Kinetic Energy Solver – kine
• Potential Energy Solver – poten
• Specific Energy Solver – specy
• Volumetric Energy Solver – volecy
• Inclined Plane Force Components – slopef
• Speed-Distance-Time Solver – vst
• Uniformly Accelerated Motion – accmotion
• Circular Motion Speed – circvel
• Orbital Speed from Arc Angle – orbvel
• Free-Fall Solver – freefall
• Quick Reference

Overview

The Physics panel (specifically the Force and Work group) provides named-parameter solvers for common physics relations. All functions accept named parameters and return a scalar or named result. Use the = display operator to see the computed value with its unit (e.g. centforce(m=2, v=10, r=5) =).

Centrifugal Force Solver – centforce

Syntax

centforce(p1=v1, p2=v2, p3=v3)

Description

Solves the centrifugal force relation F = m·v² / r. Provide any three of the four quantities:

• F – Centrifugal force [N]
• m – Mass [kg]
• v – Velocity (tangential speed) [m/s]
• r – Radius of circular path [m]

The function automatically computes the missing (fourth) quantity. All parameter names are case-sensitive.

Formula

Centrifugal Force (from the rotating frame):

F = m · v² / r  [N]

or equivalently, in terms of angular velocity:

F = m · ω² · r  (where ω = v / r)

Parameters and Units

Constraint

Each parameter may appear only once in the function call. At least three parameters must be provided (exactly three).

Examples

Example 1: Compute Force from Mass, Velocity, and Radius

centforce(m=2, v=10, r=5) =

An object of mass 2 kg moving at 10 m/s in a circle of radius 5 m experiences a centrifugal force of 40 N outward.

Example 2: Compute Velocity from Force, Mass, and Radius

centforce(F=40, m=2, r=5) =

A 2 kg object in a 5 m radius circle with 40 N centrifugal force must be moving at 10 m/s.

Example 3: Compute Mass from Force, Velocity, and Radius

centforce(F=100, v=5, r=2) =

For a 100 N centrifugal force at 5 m/s in a 2 m radius, the mass is 8 kg.

Example 4: Compute Radius from Force, Mass, and Velocity

centforce(F=50, m=3, v=10) =

A 3 kg object at 10 m/s with 50 N centrifugal force requires a radius of 6 m.

Common Applications

• Vehicle turning: Calculate the force needed to keep a car on a curved road at a given speed.
• Orbital mechanics: Determine conditions for circular orbits (centrifugal ↔ gravitational equilibrium).
• Rotating machinery: Estimate stress on components due to centrifugal loading.
• Amusement park rides: Calculate forces on riders during circular motion.
• Centrifuges: Design separation equipment by computing required rotation speeds or forces.

Related Physics Concepts

Centripetal vs. Centrifugal:

• Centripetal force: The real, inward force (tension, friction, gravity) that keeps an object moving in a circle. Always directed toward the center.
• Centrifugal force: An apparent outward force observed in a rotating reference frame. It is a fictitious force that arises due to the non-inertial nature of circular motion. In the inertial frame, objects want to move straight, and the centripetal force constrains them.

The solver uses the formula from the rotating frame perspective and computes the magnitude of centrifugal acceleration: ac = v² / r, then scales by mass.

Pressure-Force-Area Solver – pfa

Syntax

pfa(p1=v1, p2=v2)

Description

Solves the pressure relation p = F / A. Provide any two of the three quantities:

• p – Pressure [Pa] (Pascal)
• F – Force [N] (Newton)
• A – Area [m²] (square meter)

The function automatically computes the missing (third) quantity. All parameter names are case-insensitive (both p and P are accepted).

Formula

Pressure (force per unit area):

p = F / A  [Pa]

where 1 Pascal = 1 Newton / m² = 1 N/m².

Parameters and Units

Constraint

Each parameter may appear only once in the function call. Exactly two parameters must be provided (the third is computed).

Examples

Example 1: Compute Pressure from Force and Area

pfa(F=100, A=0.5) =

A force of 100 N applied over an area of 0.5 m² creates a pressure of 200 Pa.

Example 2: Compute Force from Pressure and Area

pfa(p=1000, A=0.02) =

A pressure of 1000 Pa acting on an area of 0.02 m² exerts a force of 20 N.

Example 3: Compute Area from Pressure and Force

pfa(p=5000, F=500) =

To exert 500 N of force at 5000 Pa pressure, an area of 0.1 m² is required.

Common Applications

• Hydraulic systems: Calculate pressure in pumps, cylinders, and accumulators.
• Pneumatic systems: Determine force outputs from pressure and piston area.
• Material science: Estimate stress (pressure) on samples under mechanical testing.
• Tire pressure: Understand the relationship between tire pressure and load distribution.
• Cutting tools: Calculate cutting pressure (force per unit contact area) in machining.
• Geological mechanics: Model stress and strain in rock and soil analysis.

Related Physics Concepts

Stress vs. Pressure:

• Pressure: Force per unit area perpendicular to a surface (fluid or gas contexts). Always acts normal to the surface and has no direction (scalar quantity).
• Stress: More general concept in solid mechanics, including normal stress (perpendicular) and shear stress (tangential). Stress = Force / Area, but the force may be applied at an angle.

The pfa solver assumes pressure acts perpendicular to the area (hydrostatic or pneumatic context).

Work-Force-Distance Solver – work

Syntax

work(p1=v1, p2=v2)

Description

Solves the work relation W = F · s (assuming force and displacement are in the same direction). Provide any two of the three quantities:

• W – Work [J] (Joule)
• F – Force [N] (Newton)
• s – Distance/displacement [m] (meter)

The function automatically computes the missing (third) quantity. All parameter names are case-insensitive (both W and w, both s and S are accepted).

Formula

Work (force × displacement, parallel case):

W = F · s  [J]

where 1 Joule = 1 Newton · meter = 1 N·m. This formula applies when force and displacement are parallel (angle = 0°). For general angles: W = F · s · cos(θ).

Parameters and Units

Constraint

Each parameter may appear only once in the function call. Exactly two parameters must be provided (the third is computed).

Examples

Example 1: Compute Work from Force and Distance

work(F=50, s=10) =

Applying a force of 50 N over a distance of 10 m produces a work of 500 J.

Example 2: Compute Force from Work and Distance

work(W=1000, s=20) =

If 1000 J of work is done over a distance of 20 m, the force applied is 50 N.

Example 3: Compute Distance from Work and Force

work(W=750, F=25) =

To do 750 J of work with a force of 25 N, a distance of 30 m is required.

Common Applications

• Mechanics: Calculate work done by friction, gravity, or applied forces.
• Energy transfer: Determine energy required to move an object against resistance.
• Power calculations: Work is the basis for computing power (P = W / t).
• Lifting and pushing: Calculate effort needed to lift or push objects over distances.
• Thermal systems: Work done by/on a gas in compression or expansion.
• Engine efficiency: Estimate output work given input force and displacement.

Related Physics Concepts

Work vs. Energy vs. Power:

• Work (W): Energy transferred by a force acting over a distance. Measured in Joules [J]. Is a scalar quantity.
• Energy (E): Capacity to do work. Exists in many forms: kinetic, potential, thermal, etc. Also measured in Joules [J].
• Power (P): Rate of energy transfer or work per unit time. P = W / t, measured in Watts [W] = [J/s].

Important caveat: The formula W = F · s assumes that force and displacement are in the same direction (angle θ = 0°). For a general angle: W = F · s · cos(θ). If force is perpendicular to displacement (θ = 90°), no work is done.

Power Solver – power

Description

Solves the power relation P = F · s / t (also called P = W / t since work W = F · s). Power is the rate at which work is done, or the rate of energy transfer. Provide any three of the four quantities:

• P – Power [W] (Watt)
• F – Force [N] (Newton)
• s – Distance/displacement [m] (meter)
• t – Time [s] (second)

The function automatically computes the missing (fourth) quantity. All parameter names are case-insensitive.

Formula

Power (rate of work or energy transfer):

P = F · s / t  [W]
    P = W / t      [W]  (since W = F · s)

Alternative form using velocity v = s / t:

P = F · v      [W]

Usage Syntax

power(P=value, F=..., s=...)  →  Computes t [s]
    power(F=value, s=..., t=...)  →  Computes P [W]
    power(P=..., F=..., t=...)    →  Computes s [m]
    power(P=..., F=..., s=...)    →  Computes t [s]
        

Examples

Example 1: Compute Power from Force, Distance, and Time

power(F=50, s=100, t=10) =

A force of 50 N moves an object 100 m in 10 seconds. The power delivered is (50 × 100) / 10 = 500 W.

Example 2: Compute Time from Power, Force, and Distance

power(P=500, F=50, s=100) =

If a force of 50 N does 500 W of power over 100 m, it takes t = F·s / P = 10 seconds.

Example 3: Compute Force from Power, Distance, and Time

power(P=1000, s=50, t=5) =

If 1000 W is delivered to move an object 50 m in 5 seconds, the force is F = P·t / s = 100 N.

Common Applications

• Mechanics: Calculate power of engines, motors, and mechanical systems.
• Electrical appliances: Power ratings (e.g., 100 W light bulb = energy per unit time).
• Sports: Estimate power output of athletes (climbing stairs, jumping, cycling).
• Transportation: Engine power, vehicle acceleration, climbing slopes.
• Thermal systems: Heat transfer rate, power dissipation in resistors (P = I²R).
• Renewable energy: Wind turbine power output, solar panel efficiency.

Related Physics Concepts

Key relationships:

• Work (W): Energy transferred by a force acting over a distance. W = F · s [J]. Power is the time rate of this work: P = W / t.
• Energy (E): Capacity to do work. Power is the rate of energy change: P = dE / dt.
• Velocity form: If an object moves at constant velocity v = s / t, then P = F · v. This is useful when you know force and speed instead of distance and time.
• Efficiency: Actual power output / input power. No real system is 100% efficient; some energy is lost to friction, heat, etc.
• Mechanical advantage: A machine with high power can do the same work in less time (or use less force at the same speed).

Kinetic Energy Solver – kine

Description

Solves the kinetic energy relation Ek = ½ · m · v². Kinetic energy is the energy an object possesses due to its motion. Provide any two of the three quantities:

• Ek – Kinetic energy [J] (Joule)
• m – Mass [kg] (kilogram)
• v – Velocity [m/s] (meter per second)

The function automatically computes the missing (third) quantity. All parameter names are case-insensitive.

Formula

Kinetic energy (energy of motion):

Ek = ½ · m · v²  [J]

This is the energy needed to accelerate an object from rest to velocity v, or equivalently, the energy released when the object comes to rest from velocity v.

Usage Syntax

kine(Ek=value, m=...)    →  Computes v [m/s]
            kine(m=value, v=...)     →  Computes Ek [J]
            kine(Ek=value, v=...)    →  Computes m [kg]

Examples

Example 1: Compute Kinetic Energy from Mass and Velocity

kine(m=2, v=10) =

An object with mass 2 kg moving at 10 m/s has kinetic energy Ek = ½ × 2 × 10² = 100 J.

Example 2: Compute Velocity from Kinetic Energy and Mass

kine(Ek=100, m=2) =

If an object with mass 2 kg has kinetic energy of 100 J, its velocity is v = √(2·Ek/m) = 10 m/s.

Example 3: Compute Mass from Kinetic Energy and Velocity

kine(Ek=50, v=5) =

An object moving at 5 m/s with kinetic energy 50 J has mass m = 2·Ek / v² = 4 kg.

Common Applications

• Collision analysis: Calculate impact energy when objects collide.
• Sports mechanics: Estimate energy of moving athletes, balls, or projectiles.
• Vehicle dynamics: Braking distance, crash safety, kinetic energy in accidents.
• Renewable energy: Wind and hydroelectric power (flowing water/air kinetic energy).
• Thermal physics: Relate molecular motion to temperature (average kinetic energy).
• Work-energy theorem: Net work done on an object equals its change in kinetic energy.

Related Physics Concepts

Key relationships:

• Potential energy (Ep): Energy due to position or configuration. The sum Ek + Ep is conserved in a closed system (conservation of energy). When an object falls, its potential energy converts to kinetic energy.
• Work-energy theorem: The net work done on an object equals its change in kinetic energy. W = ΔEk = Ek,final − Ek,initial.
• Momentum: Related to kinetic energy but different: p = m·v (linear), while Ek = ½·m·v² (quadratic in v). An object with twice the velocity has 4 times the kinetic energy but only twice the momentum.
• Relative velocity: Kinetic energy depends on reference frame (observer). A stationary object has Ek = 0 in its own frame, but may have Ek > 0 in another frame.
• Rotational kinetic energy: For a rotating object: Ek,rot = ½·I·ω² (where I is moment of inertia, ω is angular velocity). This is the rotational analog of translational kinetic energy.

Potential Energy Solver – poten

Description

Solves the gravitational potential energy relation Ep = m · g · h. Potential energy (also called "Lageenergie" in German) is the energy an object possesses due to its position or height in a gravitational field. Provide any three of the four quantities:

• Ep – Potential energy [J] (Joule)
• m – Mass [kg] (kilogram)
• g – Gravitational acceleration [m/s²] (usually 9.81 m/s² on Earth)
• h – Height [m] (meter, relative to a reference point)

The function automatically computes the missing (fourth) quantity. All parameter names are case-insensitive.

Formula

Gravitational potential energy:

Ep = m · g · h  [J]

This is the energy stored in an object due to its elevation in a gravitational field. When the object falls, this potential energy converts to kinetic energy (conservation of energy).

Usage Syntax

poten(Ep=value, m=..., g=...)     →  Computes h [m]
poten(m=value, g=..., h=...)       →  Computes Ep [J]
poten(Ep=value, m=..., h=...)      →  Computes g [m/s²]
poten(Ep=value, g=..., h=...)      →  Computes m [kg]

Examples

Example 1: Compute Potential Energy from Mass, Gravity, and Height

poten(m=2, g=9.81, h=10) =

A 2 kg object raised 10 meters in Earth's gravity (9.81 m/s²) has potential energy Ep = 2 × 9.81 × 10 = 196.2 J.

Example 2: Compute Height from Potential Energy, Mass, and Gravity

poten(Ep=500, m=10, g=9.81) =

If a 10 kg object has potential energy of 500 J on Earth, its height is h = Ep / (m·g) ≈ 5.1 m.

Example 3: Compute Mass from Potential Energy, Gravity, and Height

poten(Ep=100, g=9.81, h=2) =

An object at height 2 m with potential energy 100 J (on Earth) has mass m = Ep / (g·h) ≈ 5.1 kg.

Common Applications

• Hydroelectric power: Water held at height has potential energy that converts to kinetic energy and then electrical energy.
• Mechanical systems: Springs, clocks, and pendulums store and release potential energy.
• Roller coasters: Height and gravitational potential energy determine how fast the cars move at lower points.
• Climbing and sports: Calculate energy required to lift objects or climb hills.
• Structural engineering: Calculate load-bearing and stress analysis based on stored energy.
• Conservation of energy: Total mechanical energy = Ek + Ep = constant (in an isolated system).

Related Physics Concepts

Key relationships:

• Kinetic energy (Ek): Energy of motion. On Earth, when an object falls from height h, its potential energy converts to kinetic energy: Ek = Ep = m·g·h, so v = √(2·g·h) at impact (ignoring air resistance).
• Conservation of mechanical energy: In a system without friction, total energy is conserved: E_total = Ek + Ep = constant. At the top: E = m·g·h (all potential). At the bottom: E = ½·m·v² (all kinetic).
• Gravitational acceleration (g): Varies by location (9.81 m/s² on Earth's surface, ~3.7 m/s² on Mars, etc.). Different values of g yield different potential energies at the same height.
• Work done by gravity: When an object falls from height h₁ to h₂, work by gravity = m·g·(h₁ − h₂). This work equals the change in potential energy: ΔEp = m·g·Δh.
• Reference point: Potential energy is relative to a reference level (usually chosen as h = 0, Ep = 0). Only changes in potential energy are physically meaningful; absolute values depend on reference choice.

Specific Energy Solver – specy

Description

Solves the specific energy relation specy = E / m. Specific energy (also called "gravimetric energy density") is the amount of energy per unit mass. It is commonly used in thermodynamics, energy storage systems, fuel characterization, and materials science. Provide any two of the three quantities:

• specy – Specific energy [J/kg] (joule per kilogram, energy density by mass)
• E – Total energy [J] (joule)
• m – Mass [kg] (kilogram)

The function automatically computes the missing (third) quantity. All parameter names are case-insensitive.

Formula

Specific energy (energy per unit mass):

specy = E / m  [J/kg]

This represents how much energy is contained in a given mass of material or system. Examples include:

• Batteries/fuel cells: Energy per kilogram of battery or fuel (Wh/kg, often converted from J/kg)
• Explosives: Energy released per kilogram of explosive material
• Rocket propellants: Specific impulse and energy per unit mass
• Food/nutrition: Caloric content per kilogram (though usually reported in kcal/kg or kJ/kg)
• Flywheels: Rotational energy per unit mass

Usage Syntax

specy(specy=value, E=...)     →  Computes m [kg]
     specy(E=value, m=...)         →  Computes specy [J/kg]
     specy(specy=value, m=...)     →  Computes E [J]

Examples

Example 1: Compute Specific Energy from Total Energy and Mass

specy(E=1000, m=2) =

An energy source with total energy 1000 J distributed over 2 kg of material has specific energy specy = E / m = 1000 / 2 = 500 J/kg.

Example 2: Compute Total Energy from Specific Energy and Mass

specy(specy=500, m=5) =

A material with specific energy 500 J/kg and mass 5 kg has total energy E = specy × m = 500 × 5 = 2500 J.

Example 3: Compute Mass from Specific Energy and Total Energy

specy(specy=400, E=2000) =

Energy totaling 2000 J with specific energy 400 J/kg corresponds to mass m = E / specy = 2000 / 400 = 5 kg.

Common Applications

• Battery technology: Lithium-ion batteries have specific energy ~250 kJ/kg (≈ 70 Wh/kg); lead-acid ~150 kJ/kg
• Rocket propellants: RP-1 (kerosene) ~43 MJ/kg; Hydrogen ~143 MJ/kg
• Explosives: TNT ~4.6 MJ/kg; PETN ~5.8 MJ/kg
• Flywheels: High-speed carbon fiber flywheels can store 100+ kJ/kg
• Hydrogen fuel: ~143 MJ/kg (very high energy density compared to chemical batteries)
• Compressed air energy storage (CAES): Much lower density than chemical fuels (~0.1–1 MJ/kg depending on pressure)

Related Physics Concepts

Key relationships:

• Energy density (volumetric): Often confused with specific energy (gravimetric). Volumetric energy density = E / V [J/m³], where V is volume. Specific energy focuses on mass, volumetric density on volume.
• Efficiency and conversion: Real systems convert specific energy with losses. For example, a battery with 500 J/kg specific energy and 80% efficiency delivers only 400 J/kg to the load.
• Comparison with specific power: Specific power (power per unit mass) [W/kg] indicates how fast energy can be delivered; specific energy indicates how much total energy is available.
• SI unit alternatives: Specific energy is sometimes expressed as Wh/kg or MJ/kg. Conversion: 1 Wh/kg = 3600 J/kg; 1 MJ/kg = 1,000,000 J/kg.
• Gravimetric vs. volumetric trade-offs: Lightweight materials may have high specific energy but low energy density if they have large volume.

Volumetric Energy Solver – volecy

Description

Solves the volumetric energy density relation volecy = E / V. Volumetric energy density is the amount of energy per unit volume. It complements specific energy (gravimetric energy density) and is commonly used in applications where space constraints matter more than mass constraints. Provide any two of the three quantities:

• volecy – Volumetric energy density [J/m³] (joule per cubic meter, energy per volume)
• E – Total energy [J] (joule)
• V – Volume [m³] (cubic meter)

The function automatically computes the missing (third) quantity. All parameter names are case-insensitive.

Formula

Volumetric energy density (energy per unit volume):

volecy = E / V  [J/m³]

This represents how much energy is stored in a given volume of material or system. Volumetric energy density is particularly important when designing compact energy storage systems where physical space is limited (unlike mass-based specific energy which matters when weight is the constraint).

Usage Syntax

volecy(volecy=value, E=...)     →  Computes V [m³]
          volecy(E=value, V=...)          →  Computes volecy [J/m³]
          volecy(volecy=value, V=...)     →  Computes E [J]

Examples

Example 1: Compute Volumetric Energy Density from Total Energy and Volume

volecy(E=1000, V=0.5) =

An energy source with total energy 1000 J stored in volume 0.5 m³ has volumetric energy density volecy = E / V = 1000 / 0.5 = 2000 J/m³.

Example 2: Compute Total Energy from Volumetric Energy Density and Volume

volecy(volecy=2000, V=2) =

A material with volumetric energy density 2000 J/m³ and volume 2 m³ has total energy E = volecy × V = 2000 × 2 = 4000 J.

Example 3: Compute Volume from Volumetric Energy Density and Total Energy

volecy(volecy=5000, E=10000) =

Energy totaling 10000 J with volumetric energy density 5000 J/m³ requires volume V = E / volecy = 10000 / 5000 = 2 m³.

Common Applications

• Battery packs (space-constrained): Lithium-ion ~1–3 MJ/m³; lead-acid ~0.5–1 MJ/m³
• Compressed gas tanks: Compressed air ~0.1–1 MJ/m³ (depending on pressure)
• Thermal energy storage: Molten salt systems ~0.5–2 GJ/m³; water ~4 MJ/m³ per °C temperature rise
• Hydrogen (gas vs. liquid): Gas at STP ~0.01 MJ/m³; liquid ~8 GJ/m³
• Flywheel systems: High-speed steel or composite flywheels ~1–10 MJ/m³
• Supercapacitors: ~0.01–0.1 MJ/m³ (lower than batteries but faster charge/discharge)
• Fuel tanks: Gasoline ~32 MJ/m³; diesel ~37 MJ/m³; natural gas (liquefied) ~22 GJ/m³

Relationship to Specific Energy (specy)

Specific energy and volumetric energy are related but complementary:

• Specific energy (specy): Energy per unit mass [J/kg]. Focus: Weight. Applications where lightweight is critical (aircraft, satellites, electric vehicles).
• Volumetric energy (volecy): Energy per unit volume [J/m³]. Focus: Space. Applications where compact size is critical (portable devices, grid storage, vehicles with fixed dimensions).
• Density trade-off: A material with high specy but low density has low volecy; a dense material with moderate specy can have higher volecy. Example: Hydrogen gas has excellent specy (143 MJ/kg) but poor volecy due to low density (~0.01 MJ/m³ at STP).
• Conversion relationship: volecy = specy × ρ, where ρ is material density [kg/m³]. Higher density = higher volumetric energy for the same specific energy.

Related Physics Concepts

Key relationships:

• Energy density variants: Depending on context, energy density may refer to gravimetric (J/kg), volumetric (J/m³), or areal (J/m²) measures.
• Practical vs. theoretical: Theoretical energy densities assume 100% conversion efficiency. Real systems lose energy to heat, resistance, and inefficiencies. Practical values are typically 50–90% of theoretical.
• Power density: Related to but distinct from energy density. Power density [W/m³] indicates rate of energy delivery per volume; energy density [J/m³] indicates total stored energy per volume.
• SI alternatives: Large energy densities are sometimes expressed as MJ/m³ or GJ/m³. Conversion: 1 MJ/m³ = 1 million J/m³; 1 GJ/m³ = 1 billion J/m³.
• Tank/container sizing: For vehicle or spacecraft applications, volumetric energy density directly determines required tank volume for a given energy budget.

Inclined Plane Force Components – slopef

Description

Decomposes forces acting on an inclined plane (sometimes called "schiefer Ebene" or "geneigte Ebene" in German). When an object rests on an inclined plane at angle α (measured from horizontal), the weight force FG (also called "Gewichtskraft") is split into two components:

• FH – Sliding/downslope component [N] (Hangabtriebskraft, parallel to the plane)
• FN – Normal force [N] (Normalkraft, perpendicular to the plane)

Provide any one of the three forces (FG, FH, or FN) and the incline angle α. The function automatically computes all three forces.

Formula

Force decomposition on an inclined plane:

FH = FG · sin(α)      [N]  (component parallel to plane)
    FN = FG · cos(α)      [N]  (component perpendicular to plane)
    FG = FH/sin(α) = FN/cos(α)  [N]  (inverse relations)

Where:

• FG – Total weight force (Gewichtskraft) [N]
• FH – Sliding component along the plane (Hangabtrieb) [N]
• FN – Normal force perpendicular to the plane (Normalkraft) [N]
• α – Incline angle from horizontal [degrees] (0° ≤ α ≤ 90°)

Usage Syntax

slopef(FG=value, angle) =     → Returns all three forces
    slopef(FH=value, angle) =     → Computes FG and FN from FH
    slopef(FN=value, angle) =     → Computes FG and FH from FN

The force parameter name is case-insensitive. The angle must be a number in degrees (0° to 90°).

Examples

Example 1: Given Weight Force FG

slopef(FG=100, 30) =

An object weighing 100 N rests on a 30° incline. It experiences a 50 N downslope force and a 86.6 N normal force.

Example 2: Given Sliding Component FH

slopef(FH=50, 30) =

If the sliding force down the plane is 50 N and the angle is 30°, the weight is 100 N and the normal force is 86.6 N.

Example 3: Given Normal Force FN

slopef(FN=100, 45) =

On a 45° incline with a normal force of 100 N (perpendicular pressure on the surface), the weight is ~141.4 N and the sliding component is also 100 N (equal at 45°).

Common Applications

• Mechanics: Analyzing forces on ramps, slides, and inclined surfaces.
• Civil Engineering: Slope stability analysis, road grades, and safety.
• Friction problems: Determining if an object will slide (needs friction > FH).
• Pulley systems: Understanding tension when pulling objects on slopes.
• Weight analysis: Separating apparent weight (FN) from true weight (FG) on a slope.

Related Physics Concepts

Key relationships:

• Normal force (FN): Always perpendicular to the surface. It is the "apparent weight" that a scale or support surface measures. FN < FG on an incline (except at α = 0°, where FN = FG).
• Friction force: Acts opposite to FH, along the plane. For static equilibrium without sliding: F_friction ≥ FH. Maximum static friction ≈ μ_s · FN.
• Limiting angle: An object will slide if tan(α) > μ_s (static friction coefficient), i.e., when FH > μ_s · FN.
• Special cases:
  • α = 0° (horizontal): FH = 0 N, FN = FG (entire weight is normal force)
  • α = 90° (vertical): FH = FG, FN = 0 N (entire weight acts downward)
  • α = 45°: FH = FN (equal components)

Speed-Distance-Time Solver – vst

Syntax

vst(p1=v1, p2=v2)

Description

Solves the uniform-motion relation v = s / t. Provide any two groups (speed, distance, time) and the missing third group is computed.

Supported Named Parameters

Formula

v = s / t
s = v · t
t = s / v

Examples

Example 1: Compute speed in km/h

vst(skm=120, th=2) =

Example 2: Compute distance in km

vst(vkmh=90, tmin=30) =

Example 3: Compute time in seconds

vst(s=500, v=5) =

Use unit-specific parameter names when you want a specific output unit. For example, skm or th drives speed output to km/h, while s and t keep SI output (m/s).

Uniformly Accelerated Motion – accmotion

Syntax

accmotion(...=..., ...=..., ...=...)

Description

Solves the kinematic equations for uniformly accelerated motion from exactly three named inputs and returns a full result table.

Supported Named Parameters

Formulas

v = v0 + a·t
s = v0·t + 0.5·a·t²

Example

accmotion(s=100, v0=0, t=5) =

Circular Motion Speed – circvel

Syntax

circvel(...=..., ...=...)

Description

Computes tangential speed for circular motion from radius and period (time for one revolution). This is a scalar function.

Supported Named Parameters

Formula

v = 2 · π · r / T

Output Unit Rule

Returns m/s for SI-style input (r|rm + T|ts), otherwise km/h when kilometer/hour-scale parameters are used (e.g. rkm, tm, th, td).

Examples

circvel(r=10, T=5) =

circvel(rkm=1, th=1) =

Orbital Speed from Arc Angle – orbvel

Syntax

orbvel(r=..., dphi=..., ts|tm|th=...) 

Description

Computes the orbital (tangential) speed for a circular arc segment from radius, swept angle, and elapsed time. The function returns a named result table with speed in both SI and practical road units.

Parameters

Formulas

s = r · Δφ
v = s / t

In DEG mode, internally: Δφ(rad) = Δφ(deg) · π / 180.

Example

orbvel(r=10, dphi=90, ts=5) =

Free-Fall Solver – freefall

Syntax

freefall(g, h)

Description

Computes ideal free fall (without air resistance) from gravitational acceleration and fall height. The function returns a result table with:

• fall time t [s]
• impact speed v [m/s]
• impact speed v [km/h]

Parameters

Formulas

t = sqrt(2h/g)
v = sqrt(2gh)

Example

freefall(9.81, 20) =

Quick Reference

Generated for RedcrabX Physics panel documentation.