Hart's inversor uses a crossed four-link cell to connect two points by circular inversion. Add a crank that makes one point travel on a circle through the fixed inversion centre, and the other point traces a straight line. The illustration uses points halfway along the cell's bars and shows all five moving links connected to two fixed pivots.
Hart's Inversors Interactive Calculator
Change the full long and short bar lengths to explore the crossed-link mechanism. The calculator also compares the short bar with an optional 2:1 size target. That comparison is not a straightness-error calculation: a 2:1 ratio is not required for ideal inversion.
Equation Used
L and l are full pivot-to-pivot lengths, with L>l>0. A is the fixed midpoint of short bar CE; G and P are midpoints of long bars EF and CD. The input crank BG has radius R=L/4 and its fixed centre is B=(R,0). The displayed dimension margin compares an optional 2:1 target only. It does not predict trace accuracy, load capacity or manufacturing acceptance.
A crossed cell with a grounded midpoint
Two long bars CD and EF have equal length L; short bars CE and DF both have length l. The long bars cross without a pivot at their crossing. A is the midpoint of CE and is fixed to the base. G is the midpoint of EF, and P is the midpoint of CD.
The red crank BG connects G to a second fixed pivot B. Its radius equals AB, so the circle traced by G passes through A. The complete straight-line mechanism has five moving links and the fixed frame, with seven revolute joints. P is a marked tracing point on CD, not a separate sliding joint.
Kempe’s 1877 lecture explains Hart’s crossed cell using corresponding division points on its bars. The midpoint version shown here is one particular choice of those points. The A-frame is another Hart straight-line arrangement; its connectivity differs from this crossed cell.
Define the lengths before calculating the constant
For the midpoint construction and full bar lengths used here, K=(L²−l²)/4. The points A, G and P are collinear and satisfy |AG|×|AP|=K on this branch. If half-bar lengths were used instead, the same constant would be written as their squared difference. The length definition matters.
Set A=(0,0), B=(R,0) and G=(R+R cos θ,R sin θ), with R=L/4. Circle closure determines E at distances l/2 from A and L/2 from G. Then C=−E and F=2G−E. D is the crossed-branch intersection of a circle of radius L about C and one of radius l about F. P=(C+D)/2.
Inversion gives P=K G/|G|². Substituting the input circle gives x_P=K/(2R) and y_P=[K/(2R)]tan(θ/2). Thus the path is a vertical line offset by d=K/(2R) from A. Its stroke between symmetric input limits ±θ_max is 2d tan(θ_max/2).
Example: 100 mm and 50 mm full bars
With L=100 mm and l=50 mm, K=(10,000−2,500)/4=1,875 mm². The input crank radius is R=25 mm. The straight trace is d=1,875/50=37.5 mm to the right of A.
At θ=0°, G is 50 mm from A and P is at (37.5,0) mm. At ±65°, the tracing point is about ±23.89 mm above or below the centre position. The full illustrated stroke is approximately 47.78 mm.
For the optional 2:1 comparison, a 50 mm short bar differs from 100/2 by zero. An allowance of 0.05 mm then leaves a margin of 0.05 mm. Changing the short bar to 51 mm gives a difference of 1 mm and a margin of −0.95 mm, but the ideal crossed linkage still has an inversion constant. It simply has different dimensions and a different trace offset.
Finite reach and assembly branches
The circle construction requires (L−l)/2 ≤ |AG| ≤ (L+l)/2. With R=L/4, the lower boundary limits the input angle to |θ| ≤ 2 arccos(1−l/L). A boundary can be a collinear, singular configuration.
The animation uses a symmetric oscillation whose half-angle is the smaller of 65° and 80% of that geometric limit. The 80% factor keeps this display away from its limiting pose; it is not a safety factor for manufactured hardware. When L≤l, the drawing stops and asks for a longer long bar.
Crossed bars need separate physical layers. Pin clearance, stiffness and dimensional errors affect a real trace, especially near singular configurations. This ideal calculator does not infer micrometre accuracy from a universal machining tolerance.
Understanding straight-line generation
This arrangement demonstrates how rotary joints can generate an exact mathematical line without a rail. It is useful for studying inversion, loop closure and the distinction between a geometric path and the accuracy of a manufactured mechanism.
The green stroke is a calculated locus rather than a guide. The view automatically scales the linkage to fit the screen; changing dimensions changes its reported geometry. Use Pause and the animation-position control to examine how the two crossed bars and the red input crank cooperate.
Frequently asked questions
Must the bars have a 2:1 ratio?
No. This midpoint construction allows L>l>0. The 2:1 readouts compare an optional size target, not a condition for exact inversion.
Why is the squared difference divided by four?
L and l are full bar lengths and the inversion points are midpoints. Their half-length squares differ by (L²−l²)/4.
Where are the fixed pivots?
A is halfway along short bar CE. B is the centre of input crank BG. Both are attached to the same fixed frame.
Is there a joint where the long bars cross?
No. The crossing is between separate layers. The four cell-corner pivots and the three joints at A, B and G are the seven revolute joints.
What does a negative comparison margin mean?
The short bar differs from L/2 by more than the selected allowance. It does not by itself mean the ideal linkage loses its straight-line property.
Why does the animation limit the input swing?
The crossed cell has a finite reachable range. This illustration stays within one assembly branch and avoids its collinear limiting positions.
Mechanism references
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