Question

In the gear train shown, gears 3 and 4 are integral. Gear 3 meshes with gear 2, and gear 4 meshes with gear 5, gear 2 is fixed. PROJECT OBJECTIVE determine the angular velocity of the carrier for the following input angular velocities N = 100T N - 301 N. - 701 N, = 200T 100 300 500 700 1000 1500 1800 o, in RPMC cow cw CCW CW CCW CW STEPS: 1. Put all equations needed for design and calculations 2. Put all steps of the calculations (including calculations of different constant and parameters) (If calculations are made using a code please add the code) 3. Plot of the results using any program or code (Excel, Matlab,...)

Answer 1

Before starting, a few things to be noted

• Clockwise direction is positive, and anticlockwise direction is negative
• -ve Clockwise means counter-clockwise, and -ve counter-Clockwise means clockwise
• When 2 gears mesh externally the direction of rotation changes. Hence we use a negative sign. Similarly, internally meshed gears rotate in the same direction, and hence not multiplied with a negative sign
• Concentric gears rotate in the same direction with the same angular velocity
• NATA=NBTB  where N is RPM and T is number of teeth. We follow the previously mentioned sign convention also while using this equation.

Using these assumptions and formulae, we can construct the following table

SITUATION	Revs of Gear 2	Revs of Gear 3	Revs of Gear 4	Revs of Gear 5	Revs of Carrier C
Carrier C fixed, Gear 2 rotated once CW	1	10 N =-	$-\frac{10}{3}$	10 N4 10 70 3* Ns =* 3 * 200 = 7 6	0
Carrier C fixed, Gear 2 rotated x times CW	x	$-\frac{10}{3}x$	$-\frac{10}{3}x$	$-\frac{7}{6}x$	0
The whole system is locked together and given y revolutions CW	x+y	$y-\frac{10}{3}x$	$y-\frac{10}{3}x$	$y-\frac{7}{6}x$	y

The finally obtained equations provide the angular velocity of each component along with their direction.

It is given that gear 2 is fixed. Thus from the above tables

$x+y=0$ or, $x=-y$

Putting this assumption in the expression for the angular velocity of gear 5 we get

$y-\frac{7}{6}x = y+\frac{7}{6}y = \frac{13}{6}y$

Thus, $w_5= \frac{13}{6}y$

or, $y= \frac{6}{13}w_5$

From the table, we notice that the angular velocity of the carrier is y RPM CW. Thus putting in the values of $w_5$ we can get the values of y. I used Excel for this purpose. The formula used to calculate it is =6*(cell number)/13 where cell number contains angular velocity of gear 5

angular velocity of gear 5	angular velocity of the carrier
100	46.15384615
-300	-138.4615385
500	230.7692308
-700	-323.0769231
1000	461.5384615
-1500	-692.3076923
1800	830.7692308

Graph plot obtained for the gear 5 velocities along X and carrier velocity along Y is as follows. Please note that +ve means CW and -ve means CCW

carrier velocity Carrier angular velocity Carrier carrier velocity -2000 -1500 -1000 -500 500 1000 1500 7 Gear 5 angular velocity 2000 X