Go | New | Find | Notify | Tools | Reply |
Member |
Yes thumperfbc on the non insulated ratchet style, however the ones I have seen are not cheap and usually specific to certain style wire connectors. Ideal Corp. makes a model 30-429 manual crimper much like the T&B one selogic had a link to but usually around 1/2 the cost. Lowes or a local electrical supply should have one. It will work for both insulated and non insulated connectors and for casual use it is what I prefer. | |||
|
Team Apathy |
Yes, I have at least two of those style. I've always been happy with them. | |||
|
Member |
Timely, indeed. I received an article at work just last week about the importance of good crimpers (not the cheesy ten-buck ones from Lowes). There were lots of good pictures of bad crimps. If anyone's interested, [email me and] I'll find it for you. God bless America. | |||
|
Member |
There's a ratchet style crimper for damn near any connector out there. Insulated, non-insulated, coax, BNC connectors, TNC connectors, F connectors (RG59/RG6), N connectors, molex, etc, etc. Here's a kit for insulated/non-insulated/etc/etc. https://www.amazon.com/HKS-Rat...-1325616405759&psc=1 This is the tooling in which the OEM connectors are installed. The way the connector wraps the wires, there is no better option, period. You can see an example of the connector wrapping the wire here... https://www.crimpingtools.com/...-0-25-1-5mm-awg24-16 Is the other crimper a good tool, yes. But the pull strength will not match that of a ratchet style crimper. And you can take that to the bank. A better crimp = less resistance, allows for higher current flow and lowers failure rates. Do people have different preferences, yes. Do both tools provide the same level of performance, no. I knew when I posted this the ole Klein/T&B folks would freak out. Thats why the first link was to the Klien branded one... lol | |||
|
Nullus Anxietas |
Please explain the physics behind how a greater resistance can result in higher current flow? You are if loose connections weren't the actual cause of the over-heated wires. For clarification, that's actually: W = I²R Note which term is squared. I've only been working with Ohm's Law for fifty years or so, and briefly taught it, along with some other stuff, in the U.S. Army Signal School. But I'm approaching Older Than Dirt age. Maybe I'm mis-remembering How It All Works. Let's see, shall we? Let us fill in some variables. Simple case: One load: V = 20 (const.) R = 10 I = 2 W = 40 Resistance goes up: R = 15 I = 1.33 <-- N.B.: Went down, not up W = 27 Note that power consumption (W) went down, despite the fact that, though current went down, resistance went up. This is because I². Now let's take something that more closely resembles the OP's situation: Two resistances in series. (note 1) V = 240V (const.) R1 (switch): 2 (note 2) R2 (heating element): 35 (const.) (note 3) I: 6.5 W (total): 1560 W (switch): 85 (Vdrop: 13, note 4) W (heating element): 1479 (Vdrop: 228) Now the switch/connectors resistance goes up: V = 240V (const.) R1 (switch): 4 R2 (heating element): 35 (const.) I: 6.2 W (total): 1488 W (switch): 154 (Vdrop: 25) W (heating elem): 1345 (Vdrop: 217) Class dismissed Note 1: The numbers won't all come out exactly, when calculating things different ways, because I'm rounding everything up/down for display. Note 2: I chose 2 ohms for the switch and terminals just so there'd be something to square In reality a good switch and connections should be a good deal less than that. Note 3: Superficial research suggests electric clothes dryer heating elements usually fall between 20 and 50 ohms. I split the difference. Note 4: I didn't calculate Vdrop in the single-load example because, obviously, all the voltage drop will be across the single load. Here's my Cheat Sheet, in case anybody might find it handy. With it, one can calculate any one of the four values (V, I, R, or W) from any two of the others. Ohm's Law and Wattage Calculations V =---= I * R W = I * V W = I²R W = V²/R R = W/I² I = √(W/R) V = √(W*R) R = V²/WThis message has been edited. Last edited by: ensigmatic, "America is at that awkward stage. It's too late to work within the system,,,, but too early to shoot the bastards." -- Claire Wolfe "If we let things terrify us, life will not be worth living." -- Seneca the Younger, Roman Stoic philosopher | |||
|
Left-Handed, NOT Left-Winged! |
Ensigmatic Johnson is right! The problem with a loose connection is that it might arc, and arcing can cause a lot of heat. BUT, if the current was too high due to a short in the heating element, the increased heat from the current could cause the connection to melt and THEN become loose from heat warpage. But if it's due to arcing then the damage may be localized. Checking the series resistance of the heating element vs. factory spec should rule in or out the heating element. | |||
|
Nullus Anxietas |
I don't get the reference, but thanks True, but... There doesn't have to be any arcing for poor connections or degraded switch contacts to generate more heat. They generate more heat because they have more power to dissipate. That wattage consumption has to go somewhere. (Energy can be neither created nor destroyed.) Where does it go? Heat. "America is at that awkward stage. It's too late to work within the system,,,, but too early to shoot the bastards." -- Claire Wolfe "If we let things terrify us, life will not be worth living." -- Seneca the Younger, Roman Stoic philosopher | |||
|
On the wrong side of the Mobius strip |
| |||
|
Nullus Anxietas |
Thanks! "America is at that awkward stage. It's too late to work within the system,,,, but too early to shoot the bastards." -- Claire Wolfe "If we let things terrify us, life will not be worth living." -- Seneca the Younger, Roman Stoic philosopher | |||
|
Powered by Social Strata | Page 1 2 3 |
Please Wait. Your request is being processed... |