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Lawnmower for launch controller ?

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Underdog

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I budgeted poorly on my first rocket. If I spend any more there will be nothing left for the motor.
In a pinch would this battery powered mower work as a "mid-power" launch controller?
Here's what the mower has:
  1. 2 (12v) batteries that allow the mower to generate 24 volts for mowing.
  2. A removable plastic "key" that has to be installed for the mower to operate.
  3. Plug in battery charger :).
  4. Red and green lights that confirm continuity when charging and light up green when the battery is charged.
  5. A 2-stage start switch that requires you to push one lever down before activating the mower with a second lever.
  6. Fuse
  7. Wheels for transportation (no lugging heavy batteries)
The plan (should this be acceptable)

  • Disconnect the motor and run the positive and negative wires to a 110 volt outlet receptacle added to the mower chassis.
  • Add alligator clips to a surplus 30 foot 16/3 gauge extension cord (already missing its female end)
  • Done
To launch:

  1. Alligator clips on the extension cord would be attached to the leads on the rocket ignitor
  2. Operator would walk back to the mower and plug in the extension cord to the 110 outlet
  3. Install the plastic "key",
  4. Throw the first safety lockout lever
  5. Finally push the second lever (not a momentary switch) for ignition.
Question:
  • Is 24 volts going to be an issue? The batteries are 20 amp/hour batteries.
  • Will the current be to much or pose a risk?
  • Can (should) one of the two batteries be removed?
  • Would the fuse blow every time you light up the ignitor?
With the mower out of commission there will more time for rockets.

View attachment 297017558818_silo.jpg
 

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tightwad

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I have never seen a small mower that required 2 12 volt batteries to start. Are sure the batteries are not 6v? My John Deere riding mower battery works well for my launch system. It usually last as long as I care to fly, which is usually about 6 rockets ranging from LP to MP and no need for a 120 battery charger. Am I missing something?
 

mkadams001

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I have launched a lot of rockets without a launch controller at all. Just touch the leads to the battery. No need for switches or lights or a continuity button. Just raw power direct from the source. it's almost too easy.

I have included a detailed drawing.FullSizeRender.jpg
 
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vance2loud

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I have never seen a small mower that required 2 12 volt batteries to start. Are sure the batteries are not 6v? My John Deere riding mower battery works well for my launch system. It usually last as long as I care to fly, which is usually about 6 rockets ranging from LP to MP and no need for a 120 battery charger. Am I missing something?
It's an electric lawnmower with no gasoline engine.
Think big cordless drill.

I can't see any reason why it wouldn't work.
As you described it I don't think you will have any continuity indicator.
 

tightwad

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It's an electric lawnmower with no gasoline engine.
Think big cordless drill.

I can't see any reason why it wouldn't work.
As you described it I don't think you will have any continuity indicator.
I think one of those 12 batteries would work just fine. I should have looked closer at the picture...I feel stupid about missing the electric mower part. (sigh)
 

Underdog

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It's an electric lawnmower with no gasoline engine.
Think big cordless drill.
I can't see any reason why it wouldn't work.
As you described it I don't think you will have any continuity indicator.
I missed that. Yes, no continuity indicator. Perhaps I could add one? What is the function of the continuity indicator. What continuity is the light confirming?
fse.JPG
 
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Underdog

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The circuit breaker is rated for 24V and 40A. It contains what appears to be a bimetal switch. Presumably the heat generated when the current rating is exceeded bends the bimetal into an open position. There is a 68 ohm resistor inside the circuit breaker.

The motor draws 12 amps at idle and I think it climbs to 15 amps under load.
decker.JPG
 

dhbarr

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This is more than enough power. I haven't seen a 24v system, but I've launched from 16ish and no trouble.

Drop some big shielded gator clips in and burn an igniter or two before you start any mods?
 
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Buckeye

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I have launched a lot of rockets without a launch controller at all. Just touch the leads to the battery. No need for switches or lights or a continuity button. Just raw power direct from the source. it's almost too easy.

I have included a detailed drawing.View attachment 297027
Like.
 

Underdog

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I have launched a lot of rockets without a launch controller at all. Just touch the leads to the battery. No need for switches or lights or a continuity button. Just raw power direct from the source. it's almost too easy.

I have included a detailed drawing.View attachment 297027
I can appreciate this. Any patent or copyright issue if I go this route?
 

Tonimus

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The redneck in me approves of the use of a lawn mower as a controller.
 

K'Tesh

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The redneck in me approves of the use of a lawn mower as a controller.
The Red Green in me thinks it needs duct tape.

 
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Kruegon

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[*]Finally push the second lever (not a momentary switch) for ignition.
This would be my only concern. Under the safety code, the ignition button must return to the off position when released (momentary switch) and this one does not. Otherwise, I say try it out.
 

Underdog

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A change in game plan? While scrounging up parts I found a "battery jumper starter box" in the trunk of a family member's car. The "jumper starter" is plugged in and charging now. If the box still works, there may be other options.
IMG_20160717_120236.jpg
 
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kjohnson

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I have used a jumper box like this as an emergency backup launch controller when the actual controller it was powering broke at the field. Hooked the extension leads to the big alligator clamps, and flipped the switch to on to launch.

kj
 

rstaff3

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I have launched a lot of rockets without a launch controller at all. Just touch the leads to the battery. No need for switches or lights or a continuity button. Just raw power direct from the source. it's almost too easy.

I have included a detailed drawing.View attachment 297027
Looks like my controller from the 70's :)
 

rstaff3

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A change in game plan? While scrounging up parts I found a "battery jumper starter box" in the trunk of a family member's car. The "jumper starter" is plugged in and charging now. If the box still works, there may be other options.
View attachment 297142
That's the type thing I use when I do the verrrrry occasional solo launch. Well, with a Pratt GO Box plugged in the lighter hole.
 

bobkrech

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Igniters are resistors, and batteries have an internal resistance that limit current capability. Igniters are activated in a series circuit consisting of the igniter, a battery and some wiring. The battery generates a zero load voltage and the circuit has to total series resistance that generates a load. The equation explaining that is V = R x I which is known as Ohm's Law. It is also written as I = V / R or R = V / I where V is the battery potential in volts, I is the current in amps, and R is the resistance in ohms.

Ohms law applies in all cases, including igniter circuits.

For a single battery in your lawnmower, the battery voltage is about 12 volts, more or less. (A fully charged 12 volt battery is actually 13.8 volts but that is not important for this discussion.)

The resistance of a typical hobby rocket igniter is 1.5 ohms +/- 100% (about 3/4 ohm to 3 ohm range). Most igniters have an all-fire current of 2 amps +/- 400% (range is approximately 0.5 min to 8 amps max).

Ohms law is Vb = Rt x Is where Vb is the battery voltage, Rt is the total circuit resistance and Is is the series circuit current.

The total resistance Rt is the resistance of the igniter, Ri, the internal resistance of the battery, Rb,, and the resistance of the wiring, Rw.

The equation would be Rt = Ri + Rb + Rw. We already mentioned the resistance range of the igniters. The resistance of the wiring, Rw, is typically low, ~0.5 ohms or less, often 0.1 ohms or less. The internal resistance of the battery depends on the battery type and size. In the case of your jump starter that is capable of delivering 300 Amps, Rb < Vb / Ib ~ 12/300 = 0.04 ohms or lower because I didn't include any wire resistance.

So with the jump starter you can expect to have a circuit resistance of Rt = 1.5 + 0.04 + 0.5 = 2.04 ohms or less. Rounding to 2 ohms, Is = 12/2 = 6 amps.

If the all-fire current is 2 amps, and your igniter will draw 6 amps, or 3 times the all-fire current, your igniter system will work just fine.

The example you posted used a copper head igniter. As it pulled 19 amps, the igniter resistance was less than Ri < Vb/Is = 12/19 = ~0.6 ohms. AT never listed the all-fire current for this igniter. While they were inexpensive to make, they are no longer used (one reason being they are electrically inefficient), but your battery will work just fine with one because of the high current battery provided you use low resistance wiring in your controller, typically Rw < 0.1 ohms.

If we go back and recalculate the delivered current with your controller, Rt = 0.6 + 0.04 + 0.1 ~ 0.74 ohms. A fully charged battery has a voltage of ~ 13.8 volts so I = V/R = 13.8/0.74 = 18.6 or ~ 19 amps which matches the benchtop test for copperhead.
 

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