# Math Help - Not sure exactly where to post this

1. ## Not sure exactly where to post this

I'm not sure if this would be more a geometry or algebra problem. I'm trying to figure out the process to find the answer to this problem.
I have a wheel that is 40 inches in diameter. On that wheel is a gear with 20 teeth. Each tooth actuates a magnetic sensor which in turn sends pulses to a controller. I'm trying to figure out how many pulses per second if the wheel is traveling 5, and 6, and 7 miles per hour and so on.
I'm sure there is an equation or string of equations that will give me the answer just not sure how to get there.
Please let me know if this needs to be in a different thread.
Thanks

2. Originally Posted by drelectron
I'm not sure if this would be more a geometry or algebra problem. I'm trying to figure out the process to find the answer to this problem.
I have a wheel that is 40 inches in diameter. On that wheel is a gear with 20 teeth. Each tooth actuates a magnetic sensor which in turn sends pulses to a controller. I'm trying to figure out how many pulses per second if the wheel is traveling 5, and 6, and 7 miles per hour and so on.
I'm sure there is an equation or string of equations that will give me the answer just not sure how to get there.
Please let me know if this needs to be in a different thread.
Thanks
Hi dreelectron,

This is just a matter of knowing the formula for the circumference of the circle and doing a lot of unit conversions. I'll do the case where the wheel is traveling 5 mi/hr and let you do the others.

The circumference of the circle is

$C = 2 \pi R = 2 \times 3.14 \times 20 \; in \approx 126 \; in$

so the distance between two teeth is

$\frac{126 \; in}{20} \approx 6.30 \; in$.

The wheel is moving at a speed of

$\frac{5 \; mi/hr \times 5280 \; ft/mi \times 12 \; in/ft}{3600 \; sec/hr} = 88 \; in/sec$

So the pulse rate is

$\frac{88 \; in/sec}{6.30 \; in} \approx 14.0 \; /sec}$

I'm assuming the teeth are along the edge of the wheel. If that's not the case, you need to clarify your question.[/edit]

3. Originally Posted by awkward
Hi dreelectron,

This is just a matter of knowing the formula for the circumference of the circle and doing a lot of unit conversions. I'll do the case where the wheel is traveling 5 mi/hr and let you do the others.

The circumference of the circle is

$C = 2 \pi R = 2 \times 3.14 \times 20 \; in \approx 126 \; in$

so the distance between two teeth is

$\frac{126 \; in}{20} \approx 6.30 \; in$.

The wheel is moving at a speed of

$\frac{5 \; mi/hr \times 5280 \; ft/mi \times 12 \; in/ft}{3600 \; sec/hr} = 88 \; in/sec$

So the pulse rate is

$\frac{88 \; in/sec}{6.30 \; in} \approx 14.0 \; /sec}$

I'm assuming the teeth are along the edge of the wheel. If that's not the case, you need to clarify your question.[/edit]

You're right I did leave that part out. The teeth are on a gear which is about 4 inches in diameter, which is mounted on the end of the axle. So the known value is 20 pulses per revolution.

4. OK, in that case the circumference of the small wheel is

$C = \pi D = 3.14 \times 4 \; in \approx 12.6 \; in$

and the distance between teeth is

$\frac{12.6 \; in}{20} \approx 0.628 \; in}$

The rim of the wheel, whether the large wheel or the small wheel, is still moving at 88 in/sec, so the pulse rate is

$\frac{88 \; in/sec}{0.628 \;in} \approx 140 \; /sec}$

5. Originally Posted by awkward
OK, in that case the circumference of the small wheel is

$C = \pi D = 3.14 \times 4 \; in \approx 12.6 \; in$

and the distance between teeth is

$\frac{12.6 \; in}{20} \approx 0.628 \; in}$

The rim of the wheel, whether the large wheel or the small wheel, is still moving at 88 in/sec, so the pulse rate is

$\frac{88 \; in/sec}{0.628 \;in} \approx 140 \; /sec}$
OK. Please correct me if I'm wrong. A wheel with a circumference of 126" moving at 88" per second will make 1.43 revolutions per second (126/88= 1.43). The smaller gear will make the same number of revolutions which would mean there would be 28.6 pulses per second (20*1.43= 28.6).
What am I missing???
This is why I posted here I seem to get different answers depending on who I ask.

6. Originally Posted by drelectron
OK. Please correct me if I'm wrong. A wheel with a circumference of 126" moving at 88" per second will make 1.43 revolutions per second (126/88= 1.43). The smaller gear will make the same number of revolutions which would mean there would be 28.6 pulses per second (20*1.43= 28.6).
What am I missing???
This is why I posted here I seem to get different answers depending on who I ask.
As long as you know the rotational rate of the wheel-axle-gear assembly, the pulse rate is equal to the number of teeth times the rotational rate of the wheel.

However, a wheel with a circumference of 126", moving at 88" per second, will have a rotation rate of 88/126 ≈ 0.6984 revolutions per second.

There will then be 20 × 0.6984 ≈ 13.97 pulses per second.

I didn't notice all the round-off going into this computation..

A wheel with a diameter of 40", traveling at 5 mph will rotate at:

[ 5 × (5280) × (12) / (3600) ] / [ π × 40] ≈ [ 88 ] / [ 125.66 ] ≈ 0.7003 revolutions per second.

Multiplying by 20, does give 14.006 pulses per second.

(Not much different.)

7. Originally Posted by drelectron
OK. Please correct me if I'm wrong. A wheel with a circumference of 126" moving at 88" per second will make 1.43 revolutions per second (126/88= 1.43). The smaller gear will make the same number of revolutions which would mean there would be 28.6 pulses per second (20*1.43= 28.6).
What am I missing???
This is why I posted here I seem to get different answers depending on who I ask.
Drelectron,

The confusion is about the connection between the small wheel and the large wheel. Are the two on the same axis, rotating at the same number of revolutions per second (it seems so, based on your latest post)? Or are the rims of the two wheels in contact, like one gear driving another (which is what I thought when I posted my answer)?

8. Thanks for the help awkward and everyone else. The trick was figuring out how many revs per second at any given MPH which gave me the pulse per second at that MPH.
Thanks again.