Steel wire rope of 6*19

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Loading Port:: China Main Port

Payment Terms:: TT OR LC

Min Order Qty:: -

Supply Capability:: 20000 m.t./month

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Quick Details

Steel Grade:Q195 Q235 45# 60# 65# 70# 80# 82B carbon steel
Standard:AISI, ASTM, BS, DIN, GB, JIS
Wire Gauge:15.5mm
Place of Origin:Zhejiang, China (Mainland)
Type:steel wire rope
Application:fishing
Alloy Or Not:Non-alloy
Model Number:6x19
Zinc coating:it depends on the single wire diameter
Tensile strength:it depends on the single wire diameter

Packaging & Delivery

Packaging Details:	wooden reel packing iron wooden reel packing
Delivery Detail:	it depends on the quantity of the order

Specifications

6*19 Steel wire rope
covering plastics for strands or ropes
configuration: GS3x24 GS3x19 SS6x19 SS6x24
diameter:3.0-9.6mm

Steel wire ropes

It might be widely used in the marine trawl fishing , and is also suitable for guardrail facilities such as the expressway, forest Park,etc. This products have Several advantages such as wear-resisting, long Duration , beautiful appearance. It is the improved products of the zinc coated wire, especially ues for offshore

Q: how do I wire the generator so it can spin 360 on the pole with out snapping the wires.... Thanks: You need some ring contacts and brushes that will be touching those rings that will be spinning while the generator does to complete a circuit without twisting wires. Make sure you get the right size rings and wire for the amount of current

Q: just bought a brinks indoor motion sensor switch and having problems installing. the regular light switch has 2 black wires that was connected to one end and a red to the other. the white wires in the wall are already together. the motion sensor switch has black, blue, red wires and its ground wire. the instructions says not to use the red wire for single switch so confused on how to wire them.: There should not be a independent conductor on the change; or, are you checking this for the circuit on the panelboard? Or, do you propose you're analyzing 40 5 Volts between the ungrounded (warm) conductor and the change-leg? you ought to no longer get any modern analyzing from the change-leg whilst the change is opened. If there is modern modern, the change is undesirable. in case you have become a analyzing between the independent (the grounded conductor) and floor on the easy container, that must be the consequence of a multi-twine community, the place the independent is shared between 2 opposite circuits. The independent will carry any unbalanced load between the two circuits. because of fact of this you ought to by no potential assume the independent no longer wearing any modern circulate. attempt changing the change. purchase one that doesn't grant any lower back-stab connections, as they are additionally the source of many issues. If the priority extends previous that, my suggestion, as continuously, is to call a qualified, authorized electrician.

Q: hello,i have lots of 8 gauge wire resting at my closet. And i'm very interested on doing the big 3 upgrade to my 2000 mitsubishi eclipse gt. May i see any improvement with 8 gauge wire?...i know that at least 4 gauge is recommended...but i dont want to spend money.: 4 gauge? LOL Go big or go home. 1/0 gauge or it didn't happen.

Q: I dont understand it I am trying to get a 1ohm load on a 2ohm dvc speaker but cant figure out how to tell if the wire you connect in the postive terminal on the sub in a neg wire or positive how do you tell what the gray line is on this diagram positive or negetive and how to tell. thanks: wire both + together into 1 wire. wire both - together into another wire. wire the + and - to the amplifier. that's it.

Q: I want to install a single-pole dimmer switch for a series of 4 pot lights in the same box (double gang) as a kill switch for a receptical already installed. There is a 14-3 wire (b/w/r/g) brigning power into the box from the main, and 14-2 wiring for the dimmer and 14-2 wire for the kill switch. How do I wire this? The single pole dimmer has a ground and two black wires. The switch to kill the receptical has only two brass screws on one side.: Double Pole Dimmer Switch

Q: Two current carrying wires are perpendicular to each other.Wire 1 is placed on top of wire 2.Wire 1 current direction is pointing North, wire 2 current direction is pointing east. Is there a force on either of the two wires?Does one wire spin due to the produced force?Please explain to me how to do this questionThanks: Two Current Carrying Wires

Q: Or speaker wire that has an RCA plug on one end?If you know of something that is inexpensive without me having to get another preamp, please help.(this is for a turntable preamp by the way, my preamp only has RCA output.): do not use speaker wire for the turntable preamp!! use RCA cables. speaker wire is balanced wire, both conductors are identical. RCA connectorized wire is coaxial, shielded and unbalanced. You need the shielded coaxial unbalanced wire.

Q: Can anyone give me some info or a website with info on which type of wire carries sound waves best?: Wire does not carry soundwaves, at least not unless the ends of the wire have tin cans on them, and the wire is pulled taut between them! Wires can carry electrical signals which can be used to reproduce the sounds that caused them. This is how sound comes from such as your stereo system's speakers. Maybe that's what you actually mean? If so, copper wires (insulated, of course) are what you want, and if the distance you run the wires is great, the heavier the wire (gauge) the better. If you are transmitting high power to speakers, the wires have to be capable of handling the electrical current involved.

Q: Two steel wires are stretched with the same tension. The first wire has a diameter of 6.00E-4 m and the second wire has a diameter of 8.90E-4 m. If the speed of waves traveling along the first wire is 53.2 m/s, what is the speed of waves traveling along the second wire?: Wave velocity v = √(T/μ) Same tension in both wires: T1 = T2 μ is the linear mass density v1 = 53.2 m/s v2 = ? = v1/v2 = √(T1/μ1) / √(T2/μ2) = √(T1μ2 / T2μ1) = √(T1/T2 * μ2/μ1) = √(μ2/μ1) Linear mass density is mass/length, mass is density*volume: m/L = ρV/L = ρAL/L = ρA = ρπr^2 Both wires are steel so their density is the same. == v1/v2 = √(μ2/μ1) = √(ρπr2^2 / ρπr1^2) = √(r2^2 / r1^2) = r2/r1 == v2 = v1r1/r2 = (53.2*3.00×10^-4) / 4.45×10^-4 = 35.9 m/s ---- Speed of waves along the second wire is 35.9 m/s

Q: Say you have a 5 cm current wire carrying 10 A going from left to right. Directly 1 c.m below the left end of this wire is a long wire that is perpendicular to the first wire and goes out of the page. What is the net force on the 5 cm wire?I've tried using F=ILB with the I of the first wire and the B of the second wire.: I didn't read the question so I was carefully working out the force. The wire is perpendicular to the first wire, so using the right hand rule you discover that the field it creates is PARALLEL to the first wire at this left end. The magnetic force is caused by the component which is PERPENDICULAR to the wire which is in fact zero. So there is no magnetic force at this point. As you move along the wire you get a diminishing amount of magnetism caused by the wire which is going out of the page but that field has a component which is DOWN the page. Therefore that part of the wire experiences a force which is into the page. ( take your right hand, put the thumb along the wire pointing to the right, the fingers point down the page, the palm points into the page which is then the direction of the force) I would be surprised if you were required to work out the magnitude of the force in this context. You can't use F= ILB because both the magnitude and the direction of the field varies at different points along the wire. If the perpendicular wire had been directly below the middle of the other wire there would have been no net force. If you were of a level where working out the force was appropriate you would need to set up the formula for B at various points along the wire, taking the vertical component only and integrate this over the range from 0 to 5 cm. Not a trivial mathematical task.

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