Engineering in Logos 3
Engines, mechanisms, circuits, structures, flows, flight, reactions and control — what to ask for, what the engine builds, and the numbers it computes, each checked — and CAD-style designs drawn in 3D.
Logos 3 builds models from what you say — curves, systems that evolve, mechanisms assembled from parts — and, in its experimental Live 3D panel, geometry you describe. This page is a field guide for engineers: 30 models across eight disciplines and 15 CAD-style designs drawn in 3D, each with the words that ask for it, what Logos builds, what to look at, and the numbers it computes.
How to read these#
- Ask
- The words to type into the Logos 3 conversation. Logos reads them and proposes a model; the engine checks the proposal — and refuses it, naming what is missing, rather than inventing a value — then computes it. Name your numbers and the proposal uses them; leave them out and it says which values it chose.
- The model below each
- Written out in exactly the form a reply proposes, and opened by the same engine, live on this page. Move a control and it is solved again. What a reply proposes for your own words may differ in its details.
- Computed
- The engine’s own arithmetic on that model — an eigenvalue, a value at a point, the moment a run peaks — done as this page was built. Each of these numbers is checked against its closed form in Socria’s test suite, so a figure here is a computation that has been verified, not a number somebody typed.
- Views to open
- A model offers what its structure supports: a curve its Slope and Values; a system its Path, Phase portrait and Against time; a mechanism its Mechanism; a field along a line Over space and time — position across, time up, the value as colour — and across a plane The field at the clock’s time, each with its Surface. The Inspector’s How it behaves lists a system’s fixed points with their eigenvalues and, where it applies, Lyapunov exponents; for a field, How it ran gives the checks it made on itself. See Logos 2 for the views in detail.
Engines#
Cycles, the crank that turns pressure into rotation, and the wheel that stores it.
Otto and Diesel cycle efficiency
AskPlot the ideal Otto and Diesel cycle efficiencies against compression ratio, with γ and the cutoff ratio as controls.
- What it builds
- Two curves over the compression ratio: η = 1 − r^(1−γ) for the Otto cycle and η = 1 − r^(1−γ)·(ρ^γ − 1)/(γ(ρ − 1)) for the Diesel cycle, both air-standard.
- What to look at
- Move γ and both curves move; raise the cutoff ratio ρ and the Diesel curve falls further below the Otto one at the same compression.
- Computed
- At a compression ratio of 10 the ideal Otto efficiency is 60.2% and the Diesel, with ρ = 2, 53.4%.
Piston motion in a slider-crank
AskModel a piston engine’s slider-crank: piston position against crank angle, with the crank radius and the rod length as controls, and the piston speed at 3000 rpm.
- What it builds
- The piston’s distance from the crank axis, x(θ) = r·cos θ + √(l² − r²·sin² θ), and its speed at a set engine speed — both exact expressions of the crank angle.
- What to look at
- The stroke is the curve’s height, and it is 2r whatever l is. Shorten the rod and the motion grows less like a sine: the speed peaks earlier than 90°.
- Computed
- The stroke is 0.08 m — twice the crank radius — and at 3000 rpm the piston reaches 13.1 m/s, at 75.1° of crank.
Energy in a flywheel
AskHow much energy does a 50 kg solid steel flywheel of radius 0.3 m store at different speeds?
- What it builds
- The stored energy E = ½·I·ω² with I = ½·m·r² for a solid disc, as a curve over the speed in rpm.
- What to look at
- Doubling the speed quadruples the energy; doubling the radius at the same mass does the same.
- Computed
- At 3000 rpm it holds 111 kJ; at 6000 rpm, 444 kJ — four times as much.
Mechanisms and vibration#
Things that oscillate, settle, ring or refuse to repeat — assembled from parts or written as equations of motion, and integrated.
A mass, a spring and a damper
AskModel a 1 kg mass on a 20 N/m spring with a 0.6 N·s/m damper, released from 1 m.
- What it builds
- A mechanism — one body, one spring and one damper to the wall — assembled into equations of motion and integrated. Nothing is a formula for the answer: the motion is computed.
- What to look at
- Open the phase portrait: the run spirals into a stable spiral at rest. Raise the damping past 2√(km) and the fixed point turns into a stable node — the motion stops oscillating.
- Computed
- The engine finds a stable spiral at (0, 0), eigenvalues −0.3 + 4.46i, −0.3 − 4.46i. By hand: ωₙ = √(k/m) = 4.472 rad/s and ζ = c/(2√(km)) = 0.0671, so the eigenvalues should be −ζωₙ ± iωₙ√(1−ζ²).
Two masses, two natural frequencies
AskTwo 1 kg masses in a line: the first on a 100 N/m spring to the wall, joined to the second by a 50 N/m spring. What are its natural frequencies?
- What it builds
- A two-body mechanism. With no damping its rest point is a centre in four dimensions, and the eigenvalues of the assembled system are ±iω for each mode.
- What to look at
- The Inspector’s “How it behaves” lists the eigenvalues: their imaginary parts are the two natural frequencies. Add damping with the control and they gain negative real parts.
- Computed
- The engine’s eigenvalues at rest: 0 + 13.1i, 0 + 5.41i, 0 − 5.41i, 0 − 13.1i. From the eigenvalues of M⁻¹K by hand, the natural frequencies are 5.412 and 13.07 rad/s.
A pendulum, swinging far
AskModel a damped pendulum released from 2.5 rad — not the small-angle one — with length and damping as controls.
- What it builds
- θ̈ = −(g/L)·sin θ − b·θ̇ as a two-state system, sin θ and all: the small-angle approximation would be a different, wrong, model at this amplitude.
- What to look at
- The phase portrait shows the whole story: a stable spiral hanging straight down, and saddles at ±π — the inverted positions, balanced but unstable.
- Computed
- The engine finds a saddle at (−3.14, 0), eigenvalues 2.99, −3.29; stable spiral at (0, 0), eigenvalues −0.15 + 3.13i, −0.15 − 3.13i; saddle at (3.14, 0), eigenvalues 2.99, −3.29.
A double pendulum, and why it is unpredictable
AskWhy is a double pendulum unpredictable? Model one released from 2.2 and 2.4 rad.
- What it builds
- The full equations of motion of two equal arms, integrated — no approximation, because an approximate chaotic trajectory looks exactly like a real one and is wrong.
- What to look at
- The Inspector’s “How it behaves” estimates the Lyapunov exponents: the largest is positive, so two nearly identical releases separate exponentially. That is what unpredictable means here.
- Computed
- The engine’s finite-time Lyapunov exponents: 0.988, 0.0516, −0.0647, −0.979 per second — the largest positive, so nearby releases diverge exponentially.
Resonance of a driven oscillator
AskPlot the steady-state amplitude of a driven mass–spring–damper against the driving frequency.
- What it builds
- The amplitude X(ω) = F₀/√((k − mω²)² + (cω)²) — the exact steady-state response — as a curve over the drive frequency.
- What to look at
- Lower the damping and the peak grows sharper and taller; it sits just below √(k/m), at √(k/m − c²/2m²).
- Computed
- The curve peaks at ω = 4.452 rad/s with amplitude 0.374 m; the formula √(k/m − c²/2m²) puts the peak at 4.452 rad/s.
Electrical#
Circuits as differential equations: the same grammar as a mass on a spring, with the names changed.
A driven RLC loop
AskModel a series RLC circuit — 1 H, 0.5 Ω, 0.05 F — driven by a 1 V sine at 4 rad/s.
- What it builds
- The loop equation L·q̈ + R·q̇ + q/C = V(t) over the charge and the current, integrated — the same grammar as a mass on a spring with the names changed.
- What to look at
- Against time, the current settles into a steady oscillation once the transient has decayed. Slide the drive rate toward 1/√(LC) and the amplitude climbs.
- Computed
- After the transient, the engine’s run reaches a current of 0.886 A; the steady-state formula V₀/√(R² + (ωL − 1/ωC)²) gives 0.894 A. Resonance is at 1/√(LC) = 4.472 rad/s.
An RC circuit charging
AskA 10 kΩ resistor charging a 100 µF capacitor from a 5 V supply: how fast does it charge?
- What it builds
- dv/dt = (Vₛ − v)/(RC) as a one-state system, integrated from empty.
- What to look at
- Against time, the voltage rises and flattens. After one time constant τ = RC it has covered 1 − 1/e of the way.
- Computed
- After one time constant, τ = RC = 1 s, the engine’s run reads 3.161 V — 63.2% of the supply, against 1 − 1/e = 63.2%.
A Van der Pol oscillator
AskModel a Van der Pol oscillator with μ = 1 — the vacuum-tube circuit that keeps itself oscillating.
- What it builds
- ẍ − μ(1 − x²)ẋ + x = 0 as a two-state system: negative damping at small amplitude, positive at large.
- What to look at
- The phase portrait: the rest point repels — an unstable spiral — and every path, from inside or outside, winds onto the same loop, the limit cycle, of amplitude about 2.
- Computed
- The engine finds an unstable spiral at (0, 0), eigenvalues 0.5 + 0.866i, 0.5 − 0.866i, and the run’s late amplitude is 2.01.
Structures#
Beams, columns and shafts from their closed-form solutions, with the load, the material and the section as controls.
A cantilever’s deflection
AskHow far does a 2 m steel cantilever with a 100 mm square section bend under 1 kN at its tip? Show the deflected shape.
- What it builds
- The Euler–Bernoulli deflection w(x) = P·x²·(3L − x)/(6EI) along the beam, with load, modulus and second moment as controls.
- What to look at
- The tip deflection is PL³/(3EI): halve the depth of the section and I drops eightfold, and the tip goes eight times further.
- Computed
- The engine evaluates the tip at 1.6 mm; PL³/(3EI) is 1.6 mm.
A simply supported beam
AskA 6 m simply supported steel beam (I = 8 × 10⁻⁵ m⁴) with a 10 kN point load at midspan: show the deflection.
- What it builds
- The exact deflection for a central point load, symmetric about midspan: w(x) = P·m·(3L² − 4m²)/(48EI), with m the distance to the nearer support.
- What to look at
- The largest deflection is at midspan, PL³/(48EI) — a sixteenth of the cantilever’s at the same span and load.
- Computed
- At midspan the engine reads 2.81 mm; PL³/(48EI) is 2.81 mm.
Euler buckling of a column
AskPlot the Euler buckling load of a steel column against its length, for pinned, fixed and cantilevered ends.
- What it builds
- P_cr = π²EI/(KL)² as a curve over the length, with the effective-length factor K as a control (1 pinned–pinned, 0.5 fixed–fixed, 2 for a flagpole).
- What to look at
- Double the length and the load a column can carry before buckling quarters. Fix both ends and it carries four times as much.
- Computed
- A 3 m column with these properties buckles at 1.83 MN; at 6 m, 0.457 MN — a quarter.
Shear stress in a shaft
AskWhat diameter does a solid steel shaft need to carry 500 N·m without the shear stress passing 40 MPa?
- What it builds
- The torsion formula for a solid circular shaft, τ = 16T/(πd³), as a curve over the diameter.
- What to look at
- Read off where the curve crosses the stress you allow. Because τ goes as 1/d³, a little more diameter buys a lot less stress.
- Computed
- At 40 mm the engine gives 39.8 MPa; at 50 mm, 20.4 MPa.
Thermal and fluids#
Bodies cooling, tanks draining, flow through ducts and pipes.
Newton cooling
AskA 0.5 kg aluminium part at 200 °C cools in 20 °C air with h = 25 W/m²K over 0.02 m². How long until it is at 50 °C?
- What it builds
- dT/dt = −hA(T − T∞)/(mc) as a one-state system — a lumped body, one temperature throughout.
- What to look at
- Against time, the excess temperature falls by e every time constant mc/(hA).
- Computed
- The time constant mc/(hA) is 900 s. After it, the engine’s run reads 86.22 °C; the excess should have fallen to 1/e of 180 °C, which is 86.22 °C.
Heat along a rod
AskA steel rod 0.5 m long is at 20 °C. One end is put into boiling water and held at 100 °C; the other end is insulated. How does the heat travel along it over two hours?
- What it builds
- The heat equation, ∂T/∂t = α ∂²T/∂x², on a line of 101 points: Crank–Nicolson in time, the left end held at 100 °C and the right insulated — a field, stepped over space and time.
- What to look at
- Over space and time: position across, time up, temperature as colour — the warmth creeps along the rod and slows as it goes. The line marks the clock’s time; open the surface to see the same run as heights. For these ends the exact series is known, and the run says how close it comes.
- Computed
- After two hours the insulated end reads 56.6 °C; the exact series says 56.6 °C. Across the whole run the engine stays within 0.0071% of the series.
A tank draining
AskA 1 m² tank holds 2 m of water and drains through a 20 cm² hole in its floor. How long does it take to empty?
- What it builds
- Torricelli: dh/dt = −(a/A)·√(2gh), integrated until the level reaches the floor — the run stops itself there.
- What to look at
- Against time the level falls fastest at first; the last half-metre takes as long as the first 1.5 m. The run ends when the tank is empty, and says so.
- Computed
- The run stops itself when the level is within half a millimetre of the floor, at t = 314.3 s; Torricelli puts that moment at 314.2 s, and the tank empties completely at (A/a)·√(2h₀/g) = 319.3 s.
Area–Mach relation in a nozzle
AskPlot the isentropic area ratio A/A* against Mach number for air.
- What it builds
- A/A* = (1/M)·[(2/(γ+1))(1 + (γ−1)M²/2)]^((γ+1)/(2(γ−1))) — the relation every converging–diverging nozzle is designed from.
- What to look at
- The curve has its minimum, exactly 1, at M = 1: the throat. Every other area ratio has two Mach numbers, one subsonic and one supersonic.
- Computed
- The engine evaluates A/A* = 1.6875 at M = 2 (exactly 1.6875 for γ = 1.4), and the minimum 1 at M = 0.9999.
Pressure drop in a pipe
AskWhat is the pressure drop along 10 m of 50 mm pipe carrying water, as the flow speed rises? Use a friction factor of 0.02.
- What it builds
- Darcy–Weisbach, Δp = f·(L/D)·ρv²/2, as a curve over the flow speed.
- What to look at
- The loss grows as the square of the speed: twice the flow costs four times the pressure. The friction factor is a value you give — Logos does not compute it from the roughness or the Reynolds number.
- Computed
- At 2 m/s the engine gives 8 kPa; at 4 m/s, 32 kPa.
A shock forming
AskA smooth wave of velocity carries itself: faster parts catch up with slower ones. Show it steepening into a shock, with a little viscosity, on a loop.
- What it builds
- Burgers’ equation, ∂u/∂t + ∂(u²/2)/∂x = ν ∂²u/∂x², on a ring of 400 points: the transport by Rusanov’s flux, the viscosity by Crank–Nicolson.
- What to look at
- Over space and time, the wave’s leading edge sharpens into a jump and then slowly decays — the viscosity sets how thin the jump is. Nothing leaves a ring, so ∫u dx is kept, and the run says how well.
- Computed
- The steepest slope grows from 6.28 s⁻¹ at the start to 155 s⁻¹ by t = 0.3 s; without viscosity the wave would turn over at t = 1/(2πA) = 0.159 s. On the ring ∫u dx is kept to 2.2e-14 of itself.
Aerospace#
Rockets, orbits and lift — the equations, with what they assume.
The rocket equation
AskPlot the Tsiolkovsky rocket equation: Δv against mass ratio, with the specific impulse as a control.
- What it builds
- Δv = I_sp·g₀·ln(m₀/m_f) as a curve over the mass ratio.
- What to look at
- The logarithm is the whole story of rocketry: each extra km/s costs the same multiple of mass, not the same amount.
- Computed
- With a mass ratio of 10 and I_sp = 300 s, the engine gives Δv = 6774 m/s.
Orbital period and altitude
AskPlot the period of a circular Earth orbit against its altitude.
- What it builds
- Kepler’s third law, T = 2π√(a³/μ), with a = R⊕ + altitude and μ for the Earth, as a curve over the altitude in km.
- What to look at
- Low orbit is about ninety minutes; the curve reaches a day at about 35 786 km — geostationary orbit.
- Computed
- At 400 km (the space station’s height) the engine gives 92.56 minutes; at 35 786 km, 23.93 hours.
Lift against airspeed
AskA light aircraft with a 16 m² wing at C_L = 0.5: plot the lift against airspeed at sea level.
- What it builds
- L = ½·ρ·v²·S·C_L as a curve over the airspeed.
- What to look at
- The lift coefficient is an input you supply — the engine does not compute aerodynamics, so it cannot tell you what C_L a wing has. It computes what follows from it.
- Computed
- At 70 m/s the engine gives 24 kN of lift.
Chemical and process#
Rates and the reactions they drive.
Consecutive reactions A → B → C
AskModel A → B → C with k₁ = 0.5/min and k₂ = 0.2/min. When is the intermediate B at its peak?
- What it builds
- Three first-order rate equations, integrated from pure A.
- What to look at
- Against time, B rises and falls. It peaks at t* = ln(k₁/k₂)/(k₁ − k₂) — move either rate and watch the peak move.
- Computed
- The engine’s run peaks at B = 0.5429 mol/L at t = 3.05 min; ln(k₁/k₂)/(k₁ − k₂) = 3.054 min.
Arrhenius: rate against temperature
AskPlot an Arrhenius rate constant from 250 to 500 K with an activation energy of 50 kJ/mol.
- What it builds
- k(T) = A·exp(−Eₐ/(RT)) as a curve over the temperature.
- What to look at
- The rule of thumb that a reaction doubles its rate every 10 K is roughly true near room temperature for Eₐ around 50 kJ/mol — read it off the curve.
- Computed
- From 300 K to 310 K the engine’s rate constant grows by a factor of 1.91.
A reaction front
AskAn autocatalytic product makes more of itself at rate r and diffuses with D along a long tube. Seed one end — how fast does the front travel?
- What it builds
- Fisher–KPP: ∂u/∂t = D ∂²u/∂x² + r u(1 − u), on a line 100 long at 401 points — diffusion by Crank–Nicolson, the reaction by Heun in half steps either side (Strang splitting).
- What to look at
- Over space and time the front is a straight edge: it travels at a constant speed. Theory puts that speed at 2√(rD), approached from below — raise r or D and the edge leans over. The controls stop where the front would grow thinner than the grid can follow.
- Computed
- While the front — where u = ½ — crosses the middle of the tube it moves at 1.94 per unit time; theory says 2√(rD) = 2, approached from below as the front settles — the lag shrinks like (3/2)·√(D/r)/t.
Patterns from a reaction
AskTwo chemicals react, U + 2V → 3V; U is fed in and V removed, and both diffuse, V more slowly. What forms from a small disturbance?
- What it builds
- Gray–Scott on a 64 × 64 periodic plane: ∂u/∂t = Dᵤ∇²u − uv² + F(1 − u) and ∂v/∂t = Dᵥ∇²v + uv² − (F + k)v, forward Euler within its stability bound, seeded in a small square at the centre.
- What to look at
- The plane at the clock’s time, v as colour — press play to watch the pattern grow out of the seed and fill the plane. F and k decide its form: spots, stripes or waves.
- Computed
- The seed covers 1.6% of the plane; by t = 2400, v exceeds 0.1 over 39% of it — the pattern has grown out of the seed. Where nothing was seeded, u = 1 and v = 0 is a solution, and the run leaves it alone until the pattern arrives.
Control#
Feedback loops, and what tuning a gain does to where a system settles.
Position control with PD feedback
AskA 1 kg mass held at zero by a PD controller with Kp = 25 and Kd = 2. How does it settle? What changes as Kd rises?
- What it builds
- The closed loop m·ẍ = −Kp·x − Kd·ẋ as a two-state system.
- What to look at
- The phase portrait shows a stable spiral: it overshoots and rings. Drag Kd past 2√(Kp·m) = 10 and the fixed point becomes a stable node — critically damped, then overdamped, with no overshoot.
- Computed
- With K_d = 2 the engine finds a stable spiral at (0, 0), eigenvalues −1 + 4.9i, −1 − 4.9i. With K_d = 14 it finds a stable node at (0, 0), eigenvalues −2.1, −11.9.
Motor speed with PI control
AskHold a motor at 100 rad/s with a PI controller: J = 0.01 kg·m², friction b = 0.1 N·m·s, K = 0.5, Kp = 1, Ki = 5. Does it settle without error?
- What it builds
- Two states — the speed and the integral of its error — under ω̇ = (−bω + K(Kp·e + Ki·z))/J and ż = e = ω_ref − ω.
- What to look at
- Against time, the speed rises to 100 rad/s and stays: the integral term removes the steady-state error a proportional controller would leave. The fixed point is at ω = ω_ref exactly.
- Computed
- The engine finds a stable node at (100, 4), eigenvalues −4.5, −55.5; after 5 s the run reads 100 rad/s.
CAD-style designs in Live 3D#
Live 3D is an experimental panel in the Logos 3 workspace — + View → Live 3D. It has no box of its own: describe a shape in the chat and it is drawn in the panel as you type; send it and it is built, and what was built is said in the conversation. Every part keeps its identity, every size is exact, and parts can rest on one another. Give a part a material and it has a mass: density × its volume. It is a geometric preview — nothing in it is loaded, stressed or simulated — and it says so.
Below are 15 designs, from a flywheel to a truss. Each figure is the design itself, not a picture of it: the words above it are read by the panel’s own reader in your browser, and the scene they build is drawn by the panel’s own 3D view. Click a figure to turn and zoom it, and click a part for its name and size. The parts, masses and section properties listed under each are computed from the same reading.
A steel flywheel Engines
Type in the chat, with Live 3D opena steel cylinder 0.6 m in diameter and 0.1 m tall called flywheel
Read as: add a steel cylinder
| Part | Shape | Size | Mass |
|---|---|---|---|
| flywheel | cylinder | radius 0.3 m, 0.1 m tall | 222 kg (nominal density) |
- everything fits in 0.6 m × 0.1 m × 0.6 m
- mass 222.0 kg over the 1 part with a density; centre of mass at (0, 0.05, 0) m — 1 of them on nominal densities, typical values rather than measurements
Its mass is computed from its exact volume and a nominal steel density, which the panel says is a typical value. Type a measured density to replace it.
A radial engine’s layout Engines
Type in the chat, with Live 3D opena steel cylinder 0.4 m in diameter and 0.3 m tall called crankcase, then an aluminium cylinder 0.12 m in diameter and 0.35 m tall at (0.4, 0.15, 0), rotate it 90 degrees about z, then copy it 6 times around a circle of radius 0.4
Read as: add a steel cylinder · add a metal cylinder at (0.4, 0.15, 0) · copy cylinder ×6 around a circle of radius 0.4 m
| Part | Shape | Size | Mass |
|---|---|---|---|
| crankcase | cylinder | radius 0.2 m, 0.3 m tall | 295.9 kg (nominal density) |
| cylinder and 6 more alike | cylinder | radius 0.06 m, 0.35 m tall | 10.69 kg (nominal density) |
- everything fits in 0.9 m × 0.35 m × 0.924 m
- mass 370.8 kg over the 8 parts with a density; centre of mass at (−5.77e-14, 0.15, 3.59e-18) m — 8 of them on nominal densities, typical values rather than measurements
Seven cylinders pointing out from the crankcase, each turned to face outward — a layout, with masses, not a working engine: nothing here fires or turns.
A coil spring Mechanisms and vibration
Type in the chat, with Live 3D opena steel tube x = 0.05*cos(t), y = 0.05*sin(t), z = 0.008*t for t from 0 to 62.83 with tube radius 0.004 called spring
Read as: add a tube along (0.05*cos(t), 0.05*sin(t), 0.008*t)
| Part | Shape | Size | Mass |
|---|---|---|---|
| spring | tube | the curve (0.05*cos(t), 0.05*sin(t), 0.008*t), t ∈ [0, 62.83], tube radius 0.004 m | 1.252 kg (nominal density) — numerical |
- everything fits in 0.108 m × 0.511 m × 0.108 m
- mass 1.252 kg over the 1 part with a density; centre of mass at (−0.00000147, 0.251, −1.36e-9) m — 1 of them on nominal densities, typical values rather than measurements
- spring: πr² and 2πr times the curve’s length 3.173 m — exact only where the tube does not overlap itself
Ten turns of 8 mm wire, swept along the helix itself. Its length — and so its volume and mass — is measured along the curve: numerically, and said so. Its stiffness is not computed.
A ball bearing Mechanisms and vibration
Type in the chat, with Live 3D opena steel ring with outer radius 0.055 m and inner radius 0.045 m, 0.02 m thick called outer race, then a steel ring with outer radius 0.025 m and inner radius 0.015 m, 0.02 m thick called inner race at (0, 0.01, 0), then a ring of 9 steel spheres 0.02 m in diameter with radius 0.035 m around the outer race
Read as: add a steel ring · add a steel ring at (0, 0.01, 0) · add 9 steel spheres in a ring
| Part | Shape | Size | Mass |
|---|---|---|---|
| outer race | ring | radii 0.055 m and 0.045 m, 0.02 m thick | 493.2 g (nominal density) |
| inner race | ring | radii 0.025 m and 0.015 m, 0.02 m thick | 197.3 g (nominal density) |
| sphere and 8 more alike | sphere | radius 0.01 m | 32.88 g (nominal density) |
- everything fits in 0.11 m × 0.02 m × 0.11 m
- mass 986.5 g over the 11 parts with a density; centre of mass at (−6.67e-14, 0.01, −2.2e-19) m — 11 of them on nominal densities, typical values rather than measurements
- outer race: section, in the part's own x–z plane: area 3.142×10⁻³ m²; second moments about its centroid — about x (∫z² dA) 3.966×10⁻⁶ m⁴, about z (∫x² dA) 3.966×10⁻⁶ m⁴ — geometry only, no load is applied
- inner race: section, in the part's own x–z plane: area 1.257×10⁻³ m²; second moments about its centroid — about x (∫z² dA) 2.670×10⁻⁷ m⁴, about z (∫x² dA) 2.670×10⁻⁷ m⁴ — geometry only, no load is applied
Two races and nine 20 mm balls on a 35 mm pitch circle, each ball touching both races at mid-height. A geometric preview: the balls do not roll, and no load rating, contact stress or life is computed.
A V-belt pulley Mechanisms and vibration
Type in the chat, with Live 3D openan iron revolve r = 0.1 - 0.02*(1 - abs(y - 0.02)/0.02) for y from 0 to 0.04 called pulley, then paint it dark grey
Read as: add a shape turned from r = 0.1 - 0.02*(1 - abs(y - 0.02)/0.02) · make pulley dark grey
| Part | Shape | Size | Mass |
|---|---|---|---|
| pulley | revolved shape | r = 0.1 - 0.02*(1 - abs(y - 0.02)/0.02) turned about the vertical, y ∈ [0, 0.04] | 8.044 kg (nominal density) — numerical |
- everything fits in 0.2 m × 0.04 m × 0.2 m
- mass 8.044 kg over the 1 part with a density; centre of mass at (0, 0.02, 0) m — 1 of them on nominal densities, typical values rather than measurements
- pulley: narrowest radius 0.08 m where its profile has y = 0.02 (its throat); ends 0.1 m and 0.1 m
- pulley: end-to-throat area ratios (r/r_min)²: 1.563 and 1.563 — geometry only
A profile turned about its axis: 200 mm across, 40 mm wide, with a V-groove 20 mm deep. Its volume — and so its mass — is integrated along the profile numerically, and said to be; the exact value, π∫r² dy, agrees to the figures shown. The belt and its grip are not modelled.
A sprocket blank Mechanisms and vibration
Type in the chat, with Live 3D opena steel star with 24 points, radius 0.1 m, inner radius 0.088 m and 0.02 m thick called sprocket
Read as: add a steel star
| Part | Shape | Size | Mass |
|---|---|---|---|
| sprocket | star | 24 points, radii 0.1 m and 0.088 m, 0.02 m thick | 4.328 kg (nominal density) |
- everything fits in 0.2 m × 0.02 m × 0.2 m
- mass 4.328 kg over the 1 part with a density; centre of mass at (0, 0.01, 0) m — 1 of them on nominal densities, typical values rather than measurements
- sprocket: section, in the part's own x–z plane: area 0.02757 m²; second moments about its centroid — about x (∫z² dA) 6.081×10⁻⁵ m⁴, about z (∫x² dA) 6.081×10⁻⁵ m⁴ — geometry only, no load is applied
Twenty-four teeth drawn as a star: an outline with an exact area and exact second moments — but not an involute or roller-chain tooth form, which Live 3D cannot draw. Its mass is that exact area times the thickness.
A bolted flange Structures
Type in the chat, with Live 3D opena steel cylinder 0.3 m in diameter and 0.02 m tall called flange, then a ring of 8 steel cylinders 0.016 m in diameter and 0.06 m tall on the flange with radius 0.12
Read as: add a steel cylinder · add 8 steel cylinders in a ring
| Part | Shape | Size | Mass |
|---|---|---|---|
| flange | cylinder | radius 0.15 m, 0.02 m tall | 11.1 kg (nominal density) |
| cylinder and 7 more alike | cylinder | radius 0.008 m, 0.06 m tall | 94.7 g (nominal density) |
- everything fits in 1.06 m × 0.06 m × 0.3 m
- mass 11.86 kg over the 9 parts with a density; centre of mass at (0.0497, 0.0113, 0) m — 9 of them on nominal densities, typical values rather than measurements
Eight bolts on a 240 mm bolt circle, resting on the flange: move the flange and they go with it. Their preload, and the joint’s strength, are not computed.
A steel I-beam Structures
Type in the chat, with Live 3D opena steel polygon with corners (-0.1, -0.15), (0.1, -0.15), (0.1, -0.135), (0.0045, -0.135), (0.0045, 0.135), (0.1, 0.135), (0.1, 0.15), (-0.1, 0.15), (-0.1, 0.135), (-0.0045, 0.135), (-0.0045, -0.135), (-0.1, -0.135), 3 m long called beam, then rotate it 90 degrees about x
Read as: add a shape with 12 corners · rotate beam 90° about x
| Part | Shape | Size | Mass |
|---|---|---|---|
| beam | shape | 12 corners, 3 m thick | 198.5 kg (nominal density) |
- everything fits in 0.2 m × 0.3 m × 3 m
- mass 198.5 kg over the 1 part with a density; centre of mass at (−2.14e-18, 0.15, 6.56e-35) m — 1 of them on nominal densities, typical values rather than measurements
- beam: section, in the part's own x–z plane: area 8.430×10⁻³ m²; second moments about its centroid — about x (∫z² dA) 1.367×10⁻⁴ m⁴, about z (∫x² dA) 2.002×10⁻⁵ m⁴ — geometry only, no load is applied
Your own twelve corners — 200 mm flanges 15 mm thick on a 9 mm web, 300 mm deep — extruded 3 m and laid down. The section’s area and its second moments about the centroid come exactly from the corners (Green’s theorem): about its own x axis, horizontal once it is laid down, I = 1.367×10⁻⁴ m⁴ — the I the beam examples above take as a control. No load is applied and no stress is computed.
A truss, laid out Structures
Type in the chat, with Live 3D opena steel box 4 m wide, 0.1 m tall and 0.1 m deep called bottom chord, then a steel box 4 m wide, 0.1 m tall and 0.1 m deep at (0, 1.05, 0) called top chord, then a steel box 0.1 m wide, 1 m tall and 0.1 m deep at (-1.95, 0.55, 0) called post, then copy it 4 times 0.975 m to the right, then a steel box 1.4 m wide, 0.06 m tall and 0.06 m deep at (-1.4625, 0.55, 0) called brace, then rotate it 45.7 degrees about z, then copy it 3 times 0.975 m to the right
Read as: add a steel box · add a steel box at (0, 1.05, 0) · add a steel box at (-1.95, 0.55, 0) · copy post ×4 · add a steel box at (-1.4625, 0.55, 0) · rotate brace 45.7° about z · copy brace ×3
| Part | Shape | Size | Mass |
|---|---|---|---|
| bottom chord and 1 more alike | box | 4 m wide × 0.1 m tall × 0.1 m deep | 314 kg (nominal density) |
| post and 4 more alike | box | 0.1 m wide × 1 m tall × 0.1 m deep | 78.5 kg (nominal density) |
| brace and 3 more alike | box | 1.4 m wide × 0.06 m tall × 0.06 m deep | 39.56 kg (nominal density) |
- everything fits in 4 m × 1.1 m × 0.1 m
- mass 1.179 t over the 11 parts with a density; centre of mass at (1.81e-17, 0.55, 0) m — 11 of them on nominal densities, typical values rather than measurements
Two chords, five posts and four braces of solid square steel, placed by coordinates and by copies a bay apart. Its mass is the bars’ — a layout only: the joints are not pinned, nothing is loaded, and no member force is computed.
A length of pipe Thermal and fluids
Type in the chat, with Live 3D opena steel ring with outer radius 0.05 m and inner radius 0.045 m, 2 m thick
Read as: add a steel ring
| Part | Shape | Size | Mass |
|---|---|---|---|
| ring | ring | radii 0.05 m and 0.045 m, 2 m thick | 23.43 kg (nominal density) |
- everything fits in 0.1 m × 2 m × 0.1 m
- mass 23.43 kg over the 1 part with a density; centre of mass at (0, 1, 0) m — 1 of them on nominal densities, typical values rather than measurements
- ring: section, in the part's own x–z plane: area 1.492×10⁻³ m²; second moments about its centroid — about x (∫z² dA) 1.688×10⁻⁶ m⁴, about z (∫x² dA) 1.688×10⁻⁶ m⁴ — geometry only, no load is applied
An annulus extruded 2 m: the volume of metal is exact, π(R² − r²)·L, and the mass follows from it.
A finned heat sink Thermal and fluids
Type in the chat, with Live 3D openan aluminium box 0.1 m wide, 0.008 m tall and 0.08 m deep called base, then a row of 10 aluminium boxes 0.002 m wide, 0.04 m tall and 0.08 m deep on top of it with a gap of 0.008 m
Read as: add a metal box · add 10 metal boxes in a row
| Part | Shape | Size | Mass |
|---|---|---|---|
| base | box | 0.1 m wide × 0.008 m tall × 0.08 m deep | 172.8 g (nominal density) |
| box and 9 more alike | box | 0.002 m wide × 0.04 m tall × 0.08 m deep | 17.28 g (nominal density) |
- everything fits in 0.692 m × 0.04 m × 0.08 m
- mass 345.6 g over the 11 parts with a density; centre of mass at (0.298, 0.012, 0) m — 11 of them on nominal densities, typical values rather than measurements
Ten fins 2 mm thick with 8 mm between them, centred on the base they stand on. The panel gives the metal’s mass and every part’s surface area — the fins are where the area is. How much heat it sheds is not computed: that needs a heat-transfer coefficient (see Newtonian cooling above).
A vertical-axis wind turbine Thermal and fluids
Type in the chat, with Live 3D opena grey cylinder 0.1 m in diameter and 3 m tall called mast, then a blue box 0.05 m wide, 2 m tall and 0.4 m deep at (0.8, 1.7, 0) called blade, then copy it 2 times around the mast, then a dark grey box 0.8 m wide, 0.03 m tall and 0.05 m deep at (0.4, 2.3, 0) called arm, then copy it 2 times around the mast
Read as: add a grey cylinder · add a blue box at (0.8, 1.7, 0) · copy blade ×2 around a circle of radius 0.8 m · add a dark grey box at (0.4, 2.3, 0) · copy arm ×2 around a circle of radius 0.4 m
| Part | Shape | Size | Mass |
|---|---|---|---|
| mast | cylinder | radius 0.05 m, 3 m tall | — |
| blade and 2 more alike | box | 0.05 m wide × 2 m tall × 0.4 m deep | — |
| arm and 2 more alike | box | 0.8 m wide × 0.03 m tall × 0.05 m deep | — |
- everything fits in 1.41 m × 3 m × 1.63 m
Three straight blades and their arms, each copy turned about the mast to face the same way. A layout with no material given, so no mass: the rotor does not turn, and no power is computed — that needs the wind speed and a power coefficient.
A bell nozzle’s contour Aerospace
Type in the chat, with Live 3D openrevolve r = 0.15 + 0.35*(y - 0.5)^2 for y from 0 to 2 called nozzle
Read as: add a shape turned from r = 0.15 + 0.35*(y - 0.5)^2
| Part | Shape | Size | Mass |
|---|---|---|---|
| nozzle | revolved shape | r = 0.15 + 0.35*(y - 0.5)^2 turned about the vertical, y ∈ [0, 2] | — |
- everything fits in 1.88 m × 2 m × 1.88 m
- nozzle: narrowest radius 0.15 m where its profile has y = 0.5 (its throat); ends 0.237 m and 0.938 m
- nozzle: end-to-throat area ratios (r/r_min)²: 2.507 and 39.06 — geometry only
A contour turned about its axis from your own r(y). The panel finds its narrowest radius — the throat — and the area ratios to it. The solid it turns is the shape of the passage the gas fills, not the wall around it, so it is given no material; and no thrust or flow is claimed.
A wing from a NACA section Aerospace
Type in the chat, with Live 3D opena NACA 2412 aluminium wing with chord 1.5 m and span 4 m
Read as: add a metal NACA 2412 airfoil
| Part | Shape | Size | Mass |
|---|---|---|---|
| airfoil | airfoil | NACA 2412, chord 1.5 m, span 4 m | 1.987 t (nominal density) — numerical |
- everything fits in 1.5 m × 0.182 m × 4 m
- mass 1.987 t over the 1 part with a density; centre of mass at (−0.123, 0.087, 1.44e-18) m — 1 of them on nominal densities, typical values rather than measurements
- airfoil: planform area 6.000 m²; aspect ratio b²/S = 2.667; thickest 0.18 m (12% of the chord) — geometry only, no lift is computed
The section computed from the NACA formulas, extruded to the span, with its planform area and aspect ratio. Its lift is not computed — see the lift example above for what follows once C_L is given.
A three-bladed rotor Aerospace
Type in the chat, with Live 3D opena steel cylinder 0.1 m in diameter and 0.08 m tall called hub, then an aluminium box 0.6 m wide, 0.01 m tall and 0.08 m deep at (0.35, 0.04, 0) called blade, rotate it 12 degrees about x, then copy it 2 times around the hub
Read as: add a steel cylinder · add a metal box at (0.35, 0.04, 0) · copy blade ×2 around a circle of radius 0.35 m
| Part | Shape | Size | Mass |
|---|---|---|---|
| hub | cylinder | radius 0.05 m, 0.08 m tall | 4.932 kg (nominal density) |
| blade and 2 more alike | box | 0.6 m wide × 0.01 m tall × 0.08 m deep | 1.296 kg (nominal density) |
- everything fits in 1.01 m × 0.08 m × 1.17 m
- mass 8.820 kg over the 4 parts with a density; centre of mass at (0, 0.04, 0) m — 4 of them on nominal densities, typical values rather than measurements
One blade pitched 12° about its own span, then copied around the hub: each copy keeps the pitch and faces the same way about the axis. No thrust or lift is computed — the lift example above shows what follows once a lift coefficient is given.
Where it stops#
- No finite elements, no CFD. Fields are computed on a line — heat, diffusion, reaction and transport — and on a rectangle, by diffusion and reaction. Stress in a part, flow around a wing, and heat through a solid of any real shape are not: there is no mesh of a part and no Navier–Stokes solver.
- Coefficients are inputs. A friction factor, a lift coefficient, a heat-transfer coefficient: you give them, and Logos computes what follows. It does not derive them from roughness, Reynolds number or an airfoil.
- Live 3D is not CAD. No sketches with constraints, no booleans, fillets, threads or gear teeth, no assemblies with joints, no interference checks, and no import or export of CAD files. Parts can overlap and nothing will say so.
- A model is its assumptions. Each states them. A buckling load is an ideal column’s; a cycle efficiency is an air-standard cycle’s; a tank drains through a lossless orifice. Read the assumptions before the number.
For how Logos 3 itself works — the workspace, rooms and what is shared — see Logos 3.