we all know that transactions in the blockchain are completely and totally irreversible. This is true, as long as they are CONFIRMED (or mined)…but if you send a transaction, say 50k satoshi, with too low fee, in time where others are offering higher fees, and you want to recall it or cancel it, you may very well have few hours or days to do it. How?
The principle is simple, you generate new incoming address and send new transaction sending coin back to yourself, but WITH A HIGHER FEE, so it gets processed (confirmed/mined) earlier. This is the theory. There are wallets that allow it. electrum can.
Now, we are running our own node, so there must be a way how to do exactly that with Umbrel. How do we double spend coin with umbrel, RTL, or thunderhub? How can it be done?
It is tricky, but I once dove into this topic a little when 2 years ago the fees were at a total different level. These days, you could wait ages for a tx to go through, here are some thoughts which help you getting you on your research path:
@Hakuna this is all for getting the transaction to PASS. but if you simply change your mind, and want it to CANCEL (before it gets confirmed) you need to send new transaction (double spend) for the original coin, with faster fee. that way, new one will get confirmed sooner, and old one will be ignored. But how to do it in umbrel?
Some of those discussions were going in the direction of your thoughts too: use CPFP (child pays for parent) to update the destination and trump the pending tx. The method to get there is described in the articles above, and I thought some of the comments further below went into destination overwrites. But again, those were only pointers. I don’t have that one magic wand, but bump fee and CPFP should help with direcitons.
Install Ride The Lightning (RTL) and check the box to spend unconfirmed tx, then sent it to urself with a higher fee, check mempool.space to see fee recommendations
I tried with RTL but It give back the error Invalid Output Index 2 For Transaction With 2 Outputs where 2 is the Oi# matching the channel amount in the mempool.
