• The radii of the small circles add up to that of the large circle.

    Why?

  • Press the play button to fire a projectile in the direction of the arrow.

    Move the arrow and try again.

    Notice the high points of the trajectory lie on an ellipse.  If the initial velocity of the projectile is v, and gravity is g, what is the equation of the ellipse?

  • The app uses a physical model to find the largest area triangle whose vertices lie on a smooth closed curve (defined by the red control points).

    The triangle defined by the tangent lines at the vertices is displayed, along with its medians.

    What do you notice about the maximum?

  • An abacus clock.  Our previous version illustrated an idiosyncratic use of the abacus.  In this version, we conform to international standards: the value of the "five" bead is 0 when it is in the upper position, and 5 in the lower position.

  • The x coordinate of the point is the minute hand, the y coordinate is the hour hand

  • Andrew Zhao's profile was updated 8 years, 4 months ago

  • AD, BE and CF are medians of the triangle ABC.

    Triangle GHI has GH=BE, HI=CF and GI=AD.

    HL, IJ and GK are medians of GHI.

    What is the relationship between the length of these medians and the lengths of the sides of ABC?

    Can you prove this relationship geometrically?

  • You can change the orientation of the linkage by dragging point D.

    You can find the coordinates of point F by clocking on the F button.

    Show the angle of AD by pressing the base angle button.

  • The line has equation Ax+By+C=0

    You can change line parameters and observe its location relative to the coupler curve.

    You can also press teh F button to see the coordinates of the coupler point F.

  • Enter the lengths of the links and the distances BE and EF.

    BE and EF can be negative.

  • D is the midpoint of AC and E the midpoint of BD.

    What can you say about the position of F?

  • With certain values for the link lengths, this mechanism approximates a straight line.

    Experiment with different values of AB and BC then look for AD which keeps the path of point F straight.

  • Watt’s approximate straight line linkage is used in automobile suspensions.

    Drag the red points to see the mechanism move.

    What is the angle of the approximate straight line?

    Press Show for a hint.

  • Paucellier’s linkage converts rotary motion into exact straight line motion.

    Watt’s simpler linkage is only approximate.

  • A crossed over belt makes wheels rotate in opposite directions.  This mimics the action of a gear pair.

    Drag the red point to see how the involute curves mimic the motion generated by the belt.

  • We imagine two wheels connected by a belt which is wrapped in a figure 8, so that the wheels rotate in opposite directions.

    We show half the belt.  Drag the red point to see it move.

    Now imagine we cut the belt at the blue point.  You can drag the two blue points to see the involute curves.

    Press Show to turn the curves on p…[Read more]

  • For a given input force at the slider C, this app calculates the torque generated at the crank AB.

    You can observe the ratio of output torque to input force is displayed, along with geometric quantities which can be used to compute this ratio.

  • We seek the place on the earth’s surface where a vertically suspended bar (or the rings of Saturn) look the biggest.

    A machine to solve this problem consists of a mass fixed to the earth’s surface and a couple of bars attached to the mass and constrained to pass through the ends of the bar.  A spring between the bars pushes them apart.

    The…[Read more]

  • Three masses are constrained to lie on the edges of a triangle.  Equal constant force actuators (with force F) are attached to each pair of masses.  When this system finds equilibrium, the potential energy will be at a minimum.  However, the PE in each actuator is simply F times L, where L is the length of the actuator.

    Hence minimum PE is ach…[Read more]

  • This simulation finds the Fermat point as the equilibrium position of the knot in 3 pieces of string, and as the intersection of the perpendiculars to a spring loaded equilateral triangle.

    Try dragging the corners of the yellow triangle.

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