Don't know why they didn't stick to the design I have. No latch. Door opens by lifting slightly to clear a detent on the hinge then swings, nothing at all in the hatch side....And, even if I forget to open it the hatch clears it so no harm.
I just wish that on the 342 they had designed the hatch so that it is easier to get in and out of the boat when it's open......BTW, I've broken that door latch also!!!
No longer a boat owner.....previous boat - 2005 Rinker Fiesta Vee 342
I just wish that on the 342 they had designed the hatch so that it is easier to get in and out of the boat when it's open......BTW, I've broken that door latch also!!!
[EDIT: There's no way to rewire this system to move the hatch up and down. It can only go up. See modified diagrams later in this thread. This diagram is incorrect.]
I have a question about the hatch override studs used to open the engine hatch when the batteries are dead or disconnected.
In some boats, Rinker connected the two studs directly to the hatch lift actuator. The function is simple: connect a battery one way to raise the hatch, and reverse it to lower the hatch. My 2003 342 was wired this way.
Other boats, like my 2007 370, are wired totally differently. The system uses solenoids to control the current, and each stud has a [rectifier diode].
Oddly, Rinker designed this system so that the override studs can only raise the hatch. They can't lower it. Why? Is there some safety issue?
This schematic is hard to read because some of the lines are faded, but you can see the solenoids and diodes. On one side, there's a jumper wire from the stud to the solenoid (I highlighted it yellow). When you apply +12V to that stud, it throws the solenoid and raises the hatch. The other stud doesn't have this wire.
Seems like I could add a wire on the other side and have it work in both directions.
Diodes can make up a rectifier @LaRea bu not necessarily the sole purpose, in this case they are acting as a safety such that when the helm switch is activated, there is no feedback of 12vdc back to the posts from terminal 30 of the relays. Current can only flow in the direction of the arrow, back the other way hits the wall or the “I”. It also forces the current applied to the left stud not to appear at the right and vice versa.
My interpretation of operation from what I can see - press raise (switch “up”)and the left relay fires bringing +ve to the left side of M through 87 and 30 and then completes circuit via right relay 30 and 87A to ground. Press lower (switch “down”) and the right relay fires bringing +ve to the right side of M through 87 and 30 and then completes via left relay 30 and 87A to ground. But it also momentarily applies a feed to the motor before the relay kicks in which is your extra wire you see. And I am not sure why.
What doesn’t get is how it even works one way. If these relays are in their quiescent or normally off state applying power to the posts will result in a short. Put the +ve on left it will track through 30 and 87A to ground and also to the motor - but at this point it’s floating. Put -ve on right and it tracks to ground via 30 and 87A, yes it’s to the motor too but thats a load and electricity takes the easiest path and will go to 0 resistance path over that of a motor. Reverse them - same thing.
Something is missing here - in the drawing or my electrical theory...
Agreed - as drawn, the studs shouldn't work, yet they do ... in one direction. I mean, how can you complete a circuit if current can only exit from both studs? [EDIT: you can't ... the diodes are drawn incorrectly in the Rinker diagram.]
For what it's worth, the jumper wire I marked as yellow connects to the diode's anode (upstream side), not the cathode as shown in the diagram. The diode is marked 1N1183: https://www.vishay.com/docs/93492/vs-1n1183series.pdf
Agreed - as drawn, the studs shouldn't work, yet they do ... in one direction. I mean, how can you complete a circuit if current can only exit from both studs?
Keep in mind that DC flows in one direction. AC needs to flow in two directions to complete a circuit.
2008 280 Express Cruiser, 6.2MPI, B3, Pittsburgh, PA "Blue Ayes" Go Steelers!!!
But that is the point of the diodes - they don’t allow the current to get to the posts as it hits the wall as the current flow is against the diode.
Now, FWIW now makes some sense. If you put the -ve on the left stud it is tied to ground. Then +ve on right will fire the relay there first, momentarily before the diode switches on allowing flow to 30 and then 87 as the relay is on. But thats a really small switching time and I wouldn’t think the relay would pull up quicker than the diode junction.
I had the same problem. I solved it by making a connecting a portable start battery to the 12v outlet on my electrical panel. I used two 12v male plugs with a 10 amp fuse connected with about 2 feet of wire. I plug one of the 12v male connectors into the portable start battery and the other end into the 12v outlet on the electrical panel and then use the helm hatch switch to open and close the hatch. Works fine so far and I also have use of cabin lights if needed. Good luck.
That is smart. But on my boat, a 10-amp circuit from the helm would not be enough. The hatch is so heavy that it needs 20 amps directly from the port starter battery.
A 10-amp circuit from the helm could engage the solenoids, so it would work if I had a battery installed [to power the actuator]. But what I really want is the ability to raise and lower the hatch using the studs [only], with no batteries or charger connected.
I tried LaRea’s first suggestion a couple years ago, using a shore power pig tail plugged into a 110 outlet but it did not work on my 270. I’m sure the battery charger was on but the battery switch may have been off. I assumed the lift motor is 12 volt. This year I used a micro start jump pack in the terminals at the battery switch. That worked.
@icoultha - here's how mine is actually wired. On stud #2, the output side of the diode connects not to the solenoid, but directly to the battery feed.
[EDIT - corrected and reformatted this diagram, thanks @icoultha] [EDIT - diodes have opposite polarity]
The only reason I can think of is that, imagine that the diodes were not there, or shorted, and the boat batteries were depleted somewhat but still connected. If you then accidentally operated the raise/lower switch while simultaneously applying power to the studs, you would effectively be reverse coupling the battery used to power the studs with the boat battery, which would probably have a nonoptimal outcome.
After a long explanation here I hit my photo and cleared what I wrote. Grrr.
Short form - diagram has battery polarity reversed but otherwise correct in operation. Diode prevents the studs having +ve at them when the helm buttons are operated. With +ve on left stud and -ve on right, there is +ve at the motor and the coil so when you put -ve on the right stud the extra wire provides the -ve path to fire the relay that allows the motor to drive.
With +ve right and -ve left, the motor cannot operate as the left diode stops the path of the +ve through the right coil and ground bar and through the left relay contact finding the -ve line on the stud. If you add a wire to mirror what is on the right you’ll create a short circuit as you’ll bypass the diode.
Just clarify - which way opens, +ve on 1 or on 2? Which solenoid fires?
Does polarity matter - maybe, depending upon mechanical set up. As the magnetic field flows 2 different directions when you apply power, one way the field is going in one direction pulling in the core (that is attached to the contact mechanism), typically against a spring. Apply opposite polarity, the field reverses and so will tend to repel the core against an opposite spring (springs are there to make it go back home when voltage is removed otherwise it stays there). It’s unlikely in play here but many relays also have a diode across the coil to sick the back EMF when the field collapses so it both releases cleanly and prevents extraneous voltage spikes back down the line.
The two diodes are different parts -- same appearance, opposite polarity. So it's impossible for Stud 1 to feed power into the system because it runs into the diode.
So here's the corrected diagram:
The other thing I learned is that the diode on Stud 1 is damaged! It's shorted. I'm guessing it can pass enough current to throw the solenoid, but not enough to run the motor. So that's why it won't move the hatch unless the port start battery is connected.
My conclusions: 1) There's no way it can be re-wired to go up and down. 2) If I replace the shorted diode, I can probably open the hatch when there's no port starter battery connected.
So now this makes sense it’s diode 1 being biased in the other direction as I was having a hard time working out what you had.So applying +ve to 2 and -ve to 1, the hatch operates - up I assume. +ve on 1 will do nothing normally with a good diode.
When you say diode 1 is shorted you mean it has 0 ohms measured in both directions? Your assumption sounds right re the port battery if you are using stud 1 for +ve as it finds a path to operate relay 1.
I was thinking similar but you’d need decent size wires and switch for the current of the motor. Wondering about switching what controls the solenoids.
Just a thought, the trigger wire that runs from between post 2 and the diode, if this was cut and connected to a rocker switch, same as the one at the helm, and then a wire to the trigger on relay 2 (up) and one to the trigger of relay 1 (down) you should be able to select up or down. The trigger wires don't carry a lot of amps so would not be heavy gauge wire. Do you think this would work?
@Grahamu - the problem is how to raise and lower the hatch using only the override studs, when all batteries are disconnected or dead. That's not possible with the diodes in place.
That's why I'm thinking a double-pole switch. In normal operation, the hatch motor would be connected only to the solenoids. The studs would not be connected to anything, so no diodes would be needed. Flipping the switch would disconnect the solenoids and connect the motor only to the studs, which could run in either direction.
Wiring a 20-amp switch would be easy. Finding a place to mount it is harder.
Comments
2018 Cherokee 39RL Land Yacht (Sorry...)
2002 FV 342 on Lake St. Clair - Past Commodore SHC - Vessel Examiner USCGAUX
2002 FV 342 on Lake St. Clair - Past Commodore SHC - Vessel Examiner USCGAUX
I have a question about the hatch override studs used to open the engine hatch when the batteries are dead or disconnected.
Seems like I could add a wire on the other side and have it work in both directions.
My interpretation of operation from what I can see - press raise (switch “up”)and the left relay fires bringing +ve to the left side of M through 87 and 30 and then completes circuit via right relay 30 and 87A to ground. Press lower (switch “down”) and the right relay fires bringing +ve to the right side of M through 87 and 30 and then completes via left relay 30 and 87A to ground. But it also momentarily applies a feed to the motor before the relay kicks in which is your extra wire you see. And I am not sure why.
What doesn’t get is how it even works one way. If these relays are in their quiescent or normally off state applying power to the posts will result in a short. Put the +ve on left it will track through 30 and 87A to ground and also to the motor - but at this point it’s floating. Put -ve on right and it tracks to ground via 30 and 87A, yes it’s to the motor too but thats a load and electricity takes the easiest path and will go to 0 resistance path over that of a motor. Reverse them - same thing.
Something is missing here - in the drawing or my electrical theory...
Regards,
Ian
The Third “B”
Secretary, Ravena Coeymans Yacht Club
https://www.rcyachtclub.com/
For what it's worth, the jumper wire I marked as yellow connects to the diode's anode (upstream side), not the cathode as shown in the diagram. The diode is marked 1N1183: https://www.vishay.com/docs/93492/vs-1n1183series.pdf
Keep in mind that DC flows in one direction. AC needs to flow in two directions to complete a circuit.
Go Steelers!!!
Now, FWIW now makes some sense. If you put the -ve on the left stud it is tied to ground. Then +ve on right will fire the relay there first, momentarily before the diode switches on allowing flow to 30 and then 87 as the relay is on. But thats a really small switching time and I wouldn’t think the relay would pull up quicker than the diode junction.
I still don’t quite get it.
Regards,
Ian
The Third “B”
Secretary, Ravena Coeymans Yacht Club
https://www.rcyachtclub.com/
A 10-amp circuit from the helm could engage the solenoids, so it would work if I had a battery installed [to power the actuator]. But what I really want is the ability to raise and lower the hatch using the studs [only], with no batteries or charger connected.
[EDIT - corrected and reformatted this diagram, thanks @icoultha]
[EDIT - diodes have opposite polarity]
Short form - diagram has battery polarity reversed but otherwise correct in operation. Diode prevents the studs having +ve at them when the helm buttons are operated. With +ve on left stud and -ve on right, there is +ve at the motor and the coil so when you put -ve on the right stud the extra wire provides the -ve path to fire the relay that allows the motor to drive.
With +ve right and -ve left, the motor cannot operate as the left diode stops the path of the +ve through the right coil and ground bar and through the left relay contact finding the -ve line on the stud. If you add a wire to mirror what is on the right you’ll create a short circuit as you’ll bypass the diode.
Regards,
Ian
The Third “B”
Secretary, Ravena Coeymans Yacht Club
https://www.rcyachtclub.com/
Does polarity matter - maybe, depending upon mechanical set up. As the magnetic field flows 2 different directions when you apply power, one way the field is going in one direction pulling in the core (that is attached to the contact mechanism), typically against a spring. Apply opposite polarity, the field reverses and so will tend to repel the core against an opposite spring (springs are there to make it go back home when voltage is removed otherwise it stays there). It’s unlikely in play here but many relays also have a diode across the coil to sick the back EMF when the field collapses so it both releases cleanly and prevents extraneous voltage spikes back down the line.
Regards,
Ian
The Third “B”
Secretary, Ravena Coeymans Yacht Club
https://www.rcyachtclub.com/
The two diodes are different parts -- same appearance, opposite polarity. So it's impossible for Stud 1 to feed power into the system because it runs into the diode.
So here's the corrected diagram:
The other thing I learned is that the diode on Stud 1 is damaged! It's shorted. I'm guessing it can pass enough current to throw the solenoid, but not enough to run the motor. So that's why it won't move the hatch unless the port start battery is connected.
My conclusions:
1) There's no way it can be re-wired to go up and down.
2) If I replace the shorted diode, I can probably open the hatch when there's no port starter battery connected.
When you say diode 1 is shorted you mean it has 0 ohms measured in both directions? Your assumption sounds right re the port battery if you are using stud 1 for +ve as it finds a path to operate relay 1.
Regards,
Ian
The Third “B”
Secretary, Ravena Coeymans Yacht Club
https://www.rcyachtclub.com/
Regards,
Ian
The Third “B”
Secretary, Ravena Coeymans Yacht Club
https://www.rcyachtclub.com/
Regards,
Ian
The Third “B”
Secretary, Ravena Coeymans Yacht Club
https://www.rcyachtclub.com/
That's why I'm thinking a double-pole switch. In normal operation, the hatch motor would be connected only to the solenoids. The studs would not be connected to anything, so no diodes would be needed. Flipping the switch would disconnect the solenoids and connect the motor only to the studs, which could run in either direction.
Wiring a 20-amp switch would be easy. Finding a place to mount it is harder.