# Compute MPH by measuring interval between rim bolt heads (magnetic pickup)

Discussion in 'Math' started by rjwheaton, Sep 7, 2015.

1. ### rjwheaton Thread Starter New Member

Jan 26, 2014
5
0
I have a tire rim which has 5 bolts. I have a magnetic pickup which provides a very clean pulse every time a bolt head passes. The circumference of the wheel is 78.4 inches. Obviously, I get 5 pulses per wheel revolution.

I cannot figure out the math! What I basically need is to convert the interval between bolt heads to miles per hour.

Can anyone help?

Thank you very much,

Roy

2. ### MikeML AAC Fanatic!

Oct 2, 2009
5,450
1,066
An exercise in dimensional analysis:

Say you are measuring Psec/1bolt

Psec/1bolt * 5bolts/1rev * 1rev/78.4in * 12in/1ft * 5280ft/1mi * 1min/60sec * 1hr/60min = P*5*12*5280/(78.4*60*60)hr/mi

Since we wanted mi/hr, then invert and we have: (282240/316800*P) mi/hr = (49/55*P)mi/hr = (0.890909/P)mi/hr

or Speed = 49/(55*P) mph.

Transposing, let us see what P would be at 60mph: 60 = 49/(55*P)
60*55*P =49
P= 49/(60*55) = 0.01484848sec or 14.84msec

Last edited: Sep 8, 2015
rjwheaton likes this.
3. ### Wendy Moderator

Mar 24, 2008
20,766
2,536
Outer circumference of the tire sets the distance. You need the radius.

5 pulses per rotation.

4. ### MikeML AAC Fanatic!

Oct 2, 2009
5,450
1,066
Psec/1bolt * 5bolts/1rev * 1rev/78.4in *...

The OP specified that the circumference of the wheel is 78.4in. Why do we need to include the radius in the conversion?

Last edited: Sep 8, 2015
5. ### Wendy Moderator

Mar 24, 2008
20,766
2,536
Because I missed that little detail?

6. ### WBahn Moderator

Mar 31, 2012
17,743
4,795
But the circumference of the wheel is irrelevant -- it's the circumference of the tire that matters. Perhaps the TS means the tire and not the wheel, but I'm not positive about that.

7. ### djsfantasi AAC Fanatic!

Apr 11, 2010
2,805
833
A circumference of 78.4" is a diameter of 10". Much more likely to be the diameter of a wheel.
Update: Helps to use the right equation...

Last edited: Sep 8, 2015
8. ### WBahn Moderator

Mar 31, 2012
17,743
4,795
Or it's one of the new micro-cars that are becoming the snob-appeal rage.

9. ### Wendy Moderator

Mar 24, 2008
20,766
2,536
As sensors go I like it. It would also be good for odometers too.

10. ### WBahn Moderator

Mar 31, 2012
17,743
4,795
I think driveshaft sensors are probably better since you typically get several rotations of the shaft per rotation of the tire (and you can mount multiple magnets on the shaft to further improve resolution). It also tends to average out the distance traveled by the tires as you drive due to the natural behavior of the differential.

Jan 26, 2014
5
0

12. ### rjwheaton Thread Starter New Member

Jan 26, 2014
5
0
Thank you!
Not only does that answer my question, but it helps me understand how you did the calculations.

FYI, my Harley doesn't have a drive shaft (thank heavens) and the circumference of 78.4 inches computes out to a diameter of about 24.9 inches.

And yes, that will also be my odometer. The magnetic pickup is located under the swingarm, and senses the rim mounting bolts.

Roy

13. ### WBahn Moderator

Mar 31, 2012
17,743
4,795
But the question remains. Is that the circumference of the wheel, or of the tire.

In either case, it is best to make the system calibratable so that you can zero a trip meter, drive some known distance (such as on a highway with mileage markers) and then adjust the tripmeter up/down to the correct value and use that to obtain a calibration constant.

14. ### MrAl Well-Known Member

Jun 17, 2014
2,433
490
Hi,

A 25 inch "wheel" would mean a pretty darn big tire

The short answer is if there are 5 bolts per revolution and one revolution moves the car 78.4 inches, then the vehicle moves 78.4 inches for every complete revolution.

15. ### WBahn Moderator

Mar 31, 2012
17,743
4,795
I just checked a motorcycle tire website and they have tires for wheels having diameters from 8 inches all the way up to 26 inches, so who knows.

16. ### MrAl Well-Known Member

Jun 17, 2014
2,433
490
Hi,

Oh that makes sense now. I measured the tires on my car one time but i forgot what they were now. I know they are 15 inch wheels though, but the tire sizes vary quite a bit even for that one 'wheel' (rim) size.

17. ### tcmtech Well-Known Member

Nov 4, 2013
2,034
1,659
Um yea....

78.4 / 3.1415 = 24.95" diameter or a radius of ~12.47"

So where'd the 10 come from exactly?

18. ### WBahn Moderator

Mar 31, 2012
17,743
4,795
It's a simple error that messed up the units (had they been there to get messed up) -- the classic kind that I'm always harping on.

$
A \; = \; \pi r^2
\;
r \; = \; \sqrt{\frac{A}{\pi}}
\;
d \; = \; 2r \; = \; 2 \sqrt{\frac{A}{\pi}}
$

Plug in A=78.4 and pi=3.14 and you get d = 9.99, then tack on the units you want it to be and you have a diameter of 10 inches.

Plug in A = 78.4 in and pi = 3.14 and you get d = 9.99 root-inches, which just screams out "I'm WRONG!!"

But, hey, tracking units is such a waste of time, isn't it.

19. ### MrAl Well-Known Member

Jun 17, 2014
2,433
490
Hi,

That's funny i was wondering where that "10 inches" came from too. At first i thought maybe he was holding the wheel horizontally while traveling at about 0.92 times the speed of light and measuring the diameter in the same direction as travel <chuckle>

That formula does get close to 10 inches, so maybe he just used the wrong formula as his post update suggests.

20. ### tcmtech Well-Known Member

Nov 4, 2013
2,034
1,659
At least you admit to the mistake.

I had a HS math teacher that was so by the book that when the book was wrong he would still keep right on going and mark anyone's papers wrong if they used the correct formulas opposed to his wrong book formulas and believe me the books we had had a lot of basic mistakes.