Friend of mine and I were discussing this possiblity ....
He proposed (on a DC engine) that the truck wires be connected to the AC terminals of a bridge rectifier. This would ensure the stay alive capacitor would always receive the correct polarity.
To reverse the engine, a DPDT reversing switch would be installed between the capacitor and the motor. The switch could be a push button type placed under the hood and disguised as an exhaust stack or something similar. Sure, this is not a very elegant solution, but it answers the polarity problem with DC and the capacitor.
Without having actually bench tested it (yet), we are wondering what kind of charge the capacitor would attain, and if it would even be sufficient to give much (if any) stay alive. We are also wondering what the discharge level would be .... would it be the same as the current throttle level, or would it be a short burst of the maximum voltage attained by the charge voltage ?
Any thoughts ?
Mark.
¡ uʍop ǝpısdn sı ǝɹnʇɐuƃıs ʎɯ 'dlǝɥ
Yeah, it's just not going to work. That gets around the capacitor polarity problem, but everything else remains. Energy storage is the biggest - if you are running at switching speeds, when stay alive is most needed, the cap will never charge up with much energy - it can't charge to a higher voltage than the input voltage, and the amout of energy stored is directly related to the voltage. Then as David mentioned, when the track power is interrupted, the capacitor will just dump the accumulated energy as fast as it can. I don't think the loco would speed up though - if the motor is running on 6V and the cap is charged by the 6V, it won't put out more than 6V when it discharges. Butthe low total energy will mean it wouldn't keep the motor running very long. At full throttle it would charge up with plenty of power, but at high speeds most locos can simply coast over momentary dirty spots.
--Randy
Modeling the Reading Railroad in the 1950's
Visit my web site at www.readingeastpenn.com for construction updates, DCC Info, and more.
As others have mentioned, won't work.
That was kind of our thinking, but without actually trying it, we weren't sure.
So, if the cap saw a constant voltage diet varying between 0 and 6 volts, it would still only have a charged voltage (at the time of power loss) equivalent to the voltage at the time it was lost, correct ? - whether it was 2 volts or 6 volts ?
Still can't help think that even at a lower voltage, a high enough capacitance would account for something .... think three 2 farad / 5 volt super-caps in series.
May still try it just for giggles, although I'm not holding my breath for some revolutionary concept ....
gatrhumpy As others have mentioned, won't work.
Myself included ! Still, I'm one of those eccentrics that keeps thinking ....
Enough capacitance, sure, it would do something. 3x 2F caps in series is .67F total.
Problem number eleventyhundred: At low speed, say 4 volts on the rails, it would take FOREVER to charge that much capacitance.
So while it is possible to compensate for the low voltage by having a very high value capacitor, this is offset by the increased time to actually get that energy stored up. The energy stored is dependent on the square of voltage, so doubling the voltgae results in 4x the energy. For this example, 4 volts into the .67F capacitor bank is 5.36 Joules. The same capacitor charged to 12V is 48.24 Joules.
Edit: It's too early to be thinking this hard!
Well, I guess that was a bunch of wasted brain energy on my part ! Guess there won't be a "Part Trois".
Thought I would throw it out there just the same.
This is a very complex mathematical equation that uses differentials and exponential and logarithmic decay.
BUT once a cap is charged, it acts like a power source. So it will push no more current then what the circuit dictates. But the cap's voltage will drop over time as you start to drain it.
Think of it this way...when you tip a full gallon jug over, the fluid pours out initially at a very high rate. As it starts to empty it pours less and less. It's kind of like that with caps. Now the flow rate can be resistricted by resistence (like putting a smaller hole at the jug opening) making it run longer. But the total time till empty is a function of how much power is being used by the circuit (the size of the jug opening) and the total capacitence (the size of the jug). (Farads)
Don - Specializing in layout DC->DCC conversions
Modeling C&O transition era and steel industries There's Nothing Like Big Steam!
Right but then you'd need the throttle onboard the loco to control how fast is pays the energy back out when track power is lost.
Oh wait, like DCC....
The answer to this problem has been answered decades ago, add more wipers.
"The true sign of intelligence is not knowledge but imagination."-Albert Einstein
http://gearedsteam.blogspot.com/
Geared Steam The answer to this problem has been answered decades ago, add more wipers.
That's REALLY the answer, DC or DCC. Don't mask the symptoms, solve the problem.
Similar to my idea of just using a relay triggered by switch machine contacts to avoid the short in reverse loops - auto reversers wait for a fault and then fix it, but why cause a fault in the first place?
You could use a microcontroller (like an arduino) to power an H bridge and act as an automatic DPDT switch. With an array of capacitors, you could also use multiple H bridges and some PWM to control the capacitors and keep the speed constant.
At this point though, you're less running the train off of a DC current, and more charging the train via a DC current and controlling it via a microcontroller.