DIGITAL SPEEDOMETER

I want to make a digital speedometer. Concept is to measure the rpm of the wheel for say 5 sec. the revolutions will be counted be a counter. After every 5 sec the count is latched and the counter is reset. the latched count is fed to an EPROM which has the speed data fed in depending upon the wheel diameter. the output is the corresponding speed which is diplayed on 7 segment LED. Now my problem is HOW TO GENERATE THE PULSES FOR RUNNING THE COUNTER? need help on this please

Reply to
Prateek
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There are at least 3 methods slotted optical switch magnetic senors, ie a coil with a magnet passed over it

mechanical switch, say a reed relay and a rotating magnet.

put some of those words into digikey/mouser/farnell, and see what happens

martin

Reply to
martin griffith

Unless this is a school project, just go the nearest bike shop and buy a bike computer for $10. It does all that and more.

If it is a school project, a $10 bike computer will provide you with the magnet and pickup to do what you need; use the computer to verify your project speedometer. :-)

--Yan

Reply to
Captain Dondo

Reply to
Prateek

experience. The question of how to generate the pulses has many simple answers as compared to the level of knowledge you will need just to do the relevent calculations and run the displays.

Try something simpler.

Luhan

Reply to
Luhan

I think you perhaps read more or less into the post...

I for one would have little to no problems programming just about any given platform to do the calcs and to turn on the relevant GPIO and/or send the serial data to drive the display, but I would have lots and lots of problems with the physical issues surrounding the pickup.

I know; I just went through this on a machine where we eventually had to abandon counting spokes because of the noise...

Reply to
Captain Dondo

I once had a 60-slot slotted disk made, and used an optical interrupter, which drove a counter for a one second gate, and displayed RPM directly.

Is your question about how to sense the wheel's revolution? There are lots of different ways to do that, from mechanical, to magnetic, to optical. Then, just condition your sensor outputs to drive your logic.

Good Luck! Rich

Reply to
Rich Grise

On Mon, 19 Jun 2006 10:03:56 -0700, Prateek top-posted:

A reed switch is just a switch. Just provide a +5V power supply through a resistor, say, 4K7, to one terminal of the switch, ground the other terminal, and maybe put a 1 nf capacitor in parallel with the switch.

This will be normally pulled up, and when the magnet actuates the switch, you'll get a low-going pulse. You might want to condition it a little to debounce it, and maybe drive a schmitt-trigger buffer.

Good Luck! Rich p.s.: It's considered bad form to top-post on s.e.d. Thanks!

Reply to
Rich Grise

Use magnetic spokes! :-) That will also give you the benefit of having N counts per revolution - scale your counter's gate such that the number of spoke passages per count window is equal to the RPM. (or MPH, or whatever.)

Cheers! Rich

Reply to
Rich Grise

Reply to
prateek

Reply to
prateek

On Tue, 20 Jun 2006 11:56:46 -0700, prateek top-posted:

On something like a bicycle, an optical interrupter(or?) would be a bad choice, because of the dirt, as you mention.

But I had posted this before I had the magnetic spokes inspiration - it was a collaboration, actually, because I didn't think of it until after you had said that a mechanical switch hit by the spokes was too acoustically noisy.

Good Luck! Rich

Reply to
Rich Grise

Reply to
prateek

You need a resistor, say 1Kohm or 10Kohm from the 7493clock in to ground. a small capacitor from the clock to ground may be needed as you may get multiple bounces when the switch closes, so you may need a schmitt(sp?) trigger to clean up the pulses

martin

Reply to
martin griffith

That's right - you're doing it wrong. TTL inputs normally float high; applying +5V to them doesn't have any effect. You need to pull current out of the input pin to turn on the input transistor. You do this by applying +5V through a resistor, say, 2.2K to 10K - whatever you have on hand - and run that junction to the 7493 input, then ground the other side of the switch.

I don't remember if the 8493 likes a fast-rise clock - you might have to invert it, or it just might work.

Good Luck! Rich

Reply to
Rich Grise

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