Lesson 4: Acceleration Analysis and Dynamic Forces

1. Space diagram. Choose and mark a length scale, say 1 cm = 20 mm. Draw the crank mm at , then strike an arc of mm from to cut the centre-line at the piston . Measure the rod angle: below the centre-line. ✅

   

2. Velocity polygon. On a chosen velocity scale, say 1 cm = 10 mm/s, draw mm/s perpendicular to the crank. From draw the direction perpendicular to the rod, and through the horizontal the piston must move along. They meet at . ✅

   

3. Take the one number the next drawing needs. Measuring the polygon gives mm/s and mm/s. That last one is the whole reason for this step, because it fixes the rod’s angular velocity: ✅

   The rod turns slowly compared with the crank, which is why its normal acceleration will come out small in Step 2. ✅

1. Crank-pin acceleration. With the crank at constant speed, the crank pin has only a normal (centripetal) acceleration mm/s², directed from toward the centre . From the pole draw . ✅

2. Normal part of the rod. This is where Step 1 is spent. The rod’s relative-acceleration normal part is , directed from toward , and came off the velocity polygon: ✅

   Add it from . At about a tenth of it is nearly invisible on the drawing, because the rod turns slowly. Small is not the same as ignorable: it is what tips the polygon off the crank line. ✅

3. Tangential part and closure. From the end of the normal part, draw the tangential direction (perpendicular to the rod). Its magnitude is unknown, so this is a construction line, not a measured length. The piston acceleration is horizontal (the piston slides), so draw the horizontal through . Their intersection is , and it fixes both unknowns at once. ✅

4. Measure. is the piston acceleration, reading about mm/s² toward the crank, or in magnitude. Step 3 gets the same number from the formula. ✅

1. Start from the exact piston position. Project the loop onto the centre-line. The crank contributes and the rod contributes the remaining horizontal reach: ✅

2. Make the square root differentiable by hand. Since is small, , keeping the first two terms of the binomial expansion. Then use , which is what makes the second harmonic appear: ✅

3. Differentiate twice with respect to , then multiply by . Because is constant, , so no extra terms appear: ✅

   ✅

   The first term is the primary acceleration (once per revolution); the second is the secondary (twice per revolution), and it exists only because the rod has finite length. Let and it vanishes.

4. Tabulate for : ✅

   Crank
   0° (TDC)	-1.333 (peak)
   30°	-1.033
   60°	-0.333
   90°	+0.333
   120°	+0.667
   180° (BDC)	+0.667

5. Read the peaks. The largest acceleration is at top-dead-centre, , and at bottom-dead-centre it is . The two ends differ because the secondary term adds at TDC and subtracts at BDC. At , , matching the polygon. ✅

1. Set it up and fix a speed. Open the simulator (siwit.co/CSM) and set , , and RPM. A speed must be chosen before any absolute number exists: ✅

   That single number converts every result in this Application into what the chart shows. ✅

2. Predict before you look. Multiply the values already found by : ✅

   Where	This lesson	The chart should read
   TDC ()		mm/s²
   BDC ()		mm/s²
   		mm/s²

3. Mind what the readouts are called. This is where most people go wrong. The panel reports Maximum Acceleration and Minimum Acceleration, and because the top-dead-centre value is negative, the biggest acceleration in the machine appears under Minimum: ✅

   Peak-to-Peak then reads about mm/s², the distance between them. ✅

4. Read the sign as a direction. Negative means the acceleration points back toward the crank centre. At top-dead-centre the piston is being hauled to a stop and reversed, which is why the number is both negative and the largest. ✅

5. Find the twin humps. The positive side has two peaks with a shallow dip between them at bottom-dead-centre, not a single peak. Differentiating the two-harmonic result gives , so besides the dead centres there are turning points at , that is and . This is the feature students most often mistake for an error in the simulator. ✅

The chart you should be looking at, with the same three landmarks named:

6. Lengthen the rod. Increasing shrinks the secondary term, bringing the two peaks closer together and smoothing the motion. Watch Peak-to-Peak fall as you do it. ✅

1. Space diagram. Choose and mark a length scale, say 1 cm = 20 mm. Draw the ground mm, the crank mm at , then intersect an arc of mm from with an arc of mm from . The upper intersection is . Measure and . ✅

   

2. Velocity polygon. Draw mm/s perpendicular to the crank. From draw perpendicular to the coupler, from draw perpendicular to the follower, and mark where they cross. ✅

   

3. Take the two numbers the next drawing needs. Measuring gives mm/s and mm/s, so ✅

   Note how different these are. Squared, they make the follower’s normal component nearly ten times the coupler’s, which is what gives the acceleration polygon its lopsided shape. ✅

1. Crank pin. At constant crank speed, mm/s² directed from to . Draw . ✅

2. Coupler normal. Add mm/s² from , directed to . ✅

3. Follower normal. From the pole, the follower contributes mm/s², directed to . ✅

4. Tangentials close it. Two unknown magnitudes remain, and two construction lines supply them: the direction perpendicular to the coupler drawn from the end of the coupler normal, and the direction perpendicular to the follower drawn from the end of the follower normal. They intersect at . ✅

5. Measure and convert. The two tangential lengths read and mm/s². Divide each by its link length to get an angular acceleration: ✅

   The drawing gives magnitudes; the sense comes from which way each tangential vector points on it. Here the follower’s points opposite to , so is negative, which Step 3 confirms and the callout there explains. ✅

1. Start from the loop you already closed. In complex form the four-bar loop is one line, and every equation below is a derivative of it: ✅

2. Differentiate once for velocity. Each depends on time, so the chain rule brings down : ✅

3. Differentiate again for acceleration, applying the product rule to each term. Every term produces two: one from differentiating (giving ) and one from differentiating again (giving , since ). With the crank keeps only its term: ✅

   Those six terms are exactly the six vectors in the polygon of Step 2. The terms are the normals you could draw straight away; the terms are the tangentials you had to close the figure to find.

4. Split into real and imaginary parts using , and collect the two unknowns on the left. Note that multiplying by swaps sine and cosine, which is why the terms carry the opposite trig function to the terms: ✅

   Everything on the right is already known, so this is just two linear equations in two unknowns: the same solve as the velocity analysis, with a different right-hand side. ✅

5. Solve at ( rad/s), substituting , , , :

   ✅

   These match the and mm/s² measured off the polygon, divided by and . Drawing and algebra agree, which is the point of doing both. ✅

   So the follower is decelerating at this instant even though it is still driving the output forward, which the simulator’s angular-acceleration trace confirms.

1. Set it up and fix a speed. Open the simulator (siwit.co/FBL), set , , , , the assembly to open, and RPM, so rad/s and . Everything solved above was per unit crank speed, and these two factors convert it to what the simulator reports. The assembly setting matters as much as the link lengths: the crossed circuit is a different linkage with different angles and a different , so check it before comparing anything. ✅

2. Check the velocities on screen first. The and charts are tabs in the simulator. At they should read about ✅

   Divide each by and you recover the and of Step 1. ✅

3. Open the angular-acceleration tabs. Alongside , , and the simulator plots and against crank angle. Select the tab, then . (The same values are in the downloadable CSV as Alpha3 (rad/s2) and Alpha4 (rad/s2) if you would rather read exact numbers than measure off a curve.) ✅

4. Read the value at 120 degrees. Expect ✅

   Divide by to get and rad/s², matching the loop solution exactly. The signs survive the conversion, so the follower is still decelerating while turning forward. ✅

   Angular acceleration scales with the square of crank speed, so doubling the RPM quadruples every one of these numbers while leaving the per-unit values untouched. That is why the working above is done per unit crank speed. ✅

1. From the platform velocity , differentiate again:

   ✅

2. Read the two terms. The first, , is the tangential part, present only when the scissor angle is speeding up. The second, , is a centripetal-type term present even at a constant angular rate (), and it always acts downward. So a lift raised at a steady angular rate still decelerates as it nears full height. ✅

3. Actuator consequence. The actuator must supply the platform weight plus the inertia force . Near the flat position, where the mechanical advantage is poor (force analysis), this dynamic term adds appreciably to the actuator load. ✅

The scissor closes on the single scalar relation , so its acceleration is that one line differentiated twice rather than a polygon: the scaled space diagram below is the geometric reference the two derivatives rest on.

1. Open the simulator (siwit.co/SLM) and run it at a fixed actuator speed. ✅

2. Confirm that the platform acceleration is non-zero even where the angular rate is steady, and that the power trace rises where acceleration and velocity are both large. ✅

1. Apply d’Alembert to the reciprocating mass, using the piston acceleration from Application 1. The force the rod must apply to the piston to accelerate it is ; the shaking force is the equal and opposite reaction the piston applies back to the engine, which is the inertia force : ✅

   The minus sign is not bookkeeping. At top-dead-centre is at its most negative, so is at its positive peak: the piston is being hauled downward hardest exactly when it yanks the engine upward hardest. That is the force felt at the mounts. ✅

2. Split into harmonics:

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3. Balancing. The primary force can be largely cancelled by a counterweight on the crank. The secondary force runs at twice engine speed and cannot be cancelled by a simple counterweight; it is why inline-four engines use balance shafts spinning at twice crank speed. The secondary is smaller by the factor , so a longer rod also reduces it. ✅

Plotted over one revolution, the balancing argument becomes obvious. The blue primary curve rises and falls once, so anything turning with the crank can be arranged to oppose it. The green secondary curve completes two full cycles in the same revolution, so no counterweight on the crank can follow it, and only a shaft geared to twice crank speed can:

The asymmetry of the black total curve is the same one Application 1 found in the acceleration: it reaches at top-dead-centre ( and ) but only at bottom-dead-centre (), because the secondary adds at one end and subtracts at the other.