Question about GPS

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Posted on Jan 02, 2017

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SOURCE: no voice, no map display beyond 30 mile range, slow to compute

Hi,

Did you try updating your unit's firmware?

http://www8.garmin.com/products/webupdater/howtoinstall.jsp

Posted on Jan 02, 2008

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SOURCE: 60CSX altimeter problem

I fixed this problem in my 76cs by doing a factory reset.

It kept all my maps, but dumped all waypoints and tracks.

After that, the altimieter problem was fixed.

Your reset will vary, and can easily be found on google. Mine was

http://gpsinformation.us/gps60c/g76Creview.html

Complete Reset:

Press
"Enter, Page, and Power on keys" to get to the message that says, "Do
you really want to erase all user data". Before you say "YES", I would
strongly recommend you download your waypoints, tracks, and routes to a
computer, and write down your settings as they will all be erased or
reset. Of course your software version will remain the same, and

any downloaded maps will still be there.

And
then: From the Trip Computer Page, while there push the scroll key up,
right, down, & then left to get to a diagnostic screen used by
Garmin Techs. I would not recommend playing with this screen too much,
less you want to ship your unit back to Garmin for repair!

Posted on Sep 07, 2008

SOURCE: need to pick perfect amp

I suggest to use separate amps to your sub, get a 2 channel amp for your sub and 4 channel amps for your speakers.

Here are my Top 5

- Lanzar -OPTI4000D

- MA Audio- HK-4000D

- JL Audio 1000/1V2

- OPti 22.2 Power Amp

- AudioBahn-A5100T

Posted on Apr 05, 2009

SOURCE: i have 3 hifonics zeus

Hello king3132,

Three 4 ohm subs can be wired parallel for a 1.34 ohm final load or series for a 12 ohm load. If you have the channels of the amp bridged, I hope that the subs are connected in series because the Sony XM-2200GTX is NOT rated for or stable at 1.34 ohms. It's only stable down to 4 ohms.

You definitely will not get the maximum potential of the subs from that amp. They can handle 400 watts RMS each (total 1,200 watts) and the amp is only capable of 500 watts RMS into 4 ohms (less at higher impedance), so each sub could only be getting a maximum of 166.6 watts RMS.

Hope this helps.

Posted on Jul 21, 2009

Connect one voice coil to one channel of the amp. Repeat with the other voice coil. Do not bridge the amp as your only choices would be series which would be an 8 ohm final load and cuts the amp power in half or parallel wiring which results in a 2 ohm load and very few Sony amps currently made can handle being bridged to a 2 ohm load.

Aug 28, 2016 | Sony Car Audio & Video

Please see below courtesy of engineeringtoolbox.com

Electrical power is in general rated in watt (W) or horsepower (HP). A horsepower is a unit of power equal to*746 watts* or *33000 lb ft per minute* (or *550 lb ft per second*).

A watt is equal to the power produced by a current of*1 amp* across the potential difference of *1 volt*. A watt is *1/746* of *1 horsepower*.

Even if the watt is the base unit of electrical power, it is common to rate motor power in either horsepower or kilowatts.

### Power in Watts

Direct Current
Electric power supply to a direct current (DC) motor:

*PkW = ?m U I** / 1000 (1)*

*where*

*PkW =** power (kW)*

*?m** = motor efficiency*

*U** = voltage (V)*

*I** = current (A, amps)*

Electric power supply to a alternating current (AC) motor:

Single Phase*PkW* = ?m U I* PF / 1000 (1b)*

*where *

*PF = Power Factor*

Two Phase Four Wire*PkW* = ?m 2 U I* PF / 1000 (1c)*

Three Phase*PkW* = ?m 1.73 U I* PF / 1000 (1d)*

### Power in Horsepower

Horse power can be expressed as:

Electrical power is in general rated in watt (W) or horsepower (HP). A horsepower is a unit of power equal to

A watt is equal to the power produced by a current of

Even if the watt is the base unit of electrical power, it is common to rate motor power in either horsepower or kilowatts.

Electric power supply to a alternating current (AC) motor:

Single Phase

Two Phase Four Wire

Three Phase

Php =PkW/ 0.746(2)

or

Php = (?m U I / 1000) / 0.746

=?m U I / 746(2b)

where

Php =horsepower (hp)

Feb 03, 2016 | Boating

Which current are you referring to ??......

If you think of an electrical device as a piece of plumbing, voltage is the amount of water that you send down into the pipe, resistance is the pipe's relative width or narrowness, and current is the speed with which the water flows.

Power measures the water's relative difficulty or ease making its way through the pipe.

You relate all these values to one another using a common set of physics equations known as Ohm's law.

If you need to calculate electricity's current flow, you'll need to have at least two of the three values -- voltage, resistance or power -- listed above.

Calculate current flow using voltage and resistance.

According to Ohm's law, you can express electricity's current in amps as a ratio of its voltage in volts to the resistance of the device it's flowing through in ohms -- I = E/R, respectively.

For example, if you want to know the current flow of 220 V of electricity as it flows through a laptop computer with 80 ohms of resistance, you would simply plug these values into the equation as follows: I = 220/80 = 2.75 amps.

Calculate current flow using power and resistance.

Ohm's law also states that electrical current, "I," is equal to the square root of the power dissipated as it travels through the device divided by that device's resistance.

If a light bulb dissipates 80 watts of power and has a resistance of 55 ohms, you can calculate the electricity's current as follows: I = sqrt(80/55) = sqrt(1.4545) = 1.20 amps.

Calculate current flow using power and voltage.

If you have a space heater which dissipates 420 watts of power when it takes in 120 V of electricity, Ohm's law states you can calculate this electricity's current using the equation "I = P/E." For this example, compute current like so: I =420/120 = 3.5 amps.

http://www.the12volt.com/ohm/ohmslaw.asp

If you think of an electrical device as a piece of plumbing, voltage is the amount of water that you send down into the pipe, resistance is the pipe's relative width or narrowness, and current is the speed with which the water flows.

Power measures the water's relative difficulty or ease making its way through the pipe.

You relate all these values to one another using a common set of physics equations known as Ohm's law.

If you need to calculate electricity's current flow, you'll need to have at least two of the three values -- voltage, resistance or power -- listed above.

Calculate current flow using voltage and resistance.

According to Ohm's law, you can express electricity's current in amps as a ratio of its voltage in volts to the resistance of the device it's flowing through in ohms -- I = E/R, respectively.

For example, if you want to know the current flow of 220 V of electricity as it flows through a laptop computer with 80 ohms of resistance, you would simply plug these values into the equation as follows: I = 220/80 = 2.75 amps.

Calculate current flow using power and resistance.

Ohm's law also states that electrical current, "I," is equal to the square root of the power dissipated as it travels through the device divided by that device's resistance.

If a light bulb dissipates 80 watts of power and has a resistance of 55 ohms, you can calculate the electricity's current as follows: I = sqrt(80/55) = sqrt(1.4545) = 1.20 amps.

Calculate current flow using power and voltage.

If you have a space heater which dissipates 420 watts of power when it takes in 120 V of electricity, Ohm's law states you can calculate this electricity's current using the equation "I = P/E." For this example, compute current like so: I =420/120 = 3.5 amps.

http://www.the12volt.com/ohm/ohmslaw.asp

Aug 13, 2013 | Computers & Internet

you are missing the voltage supply here to get an definitive answer

for 240 volts drawing 4.8 amps the required wattage is 1152

for 120 volts the wattage is 576 watts

for 120 volts with a current of 900 watts , it equates to 7.5 amps

with 240 volts with a current of 1150 watts , it equates to 4.79 amps

so you can see the problem by not stating the supply voltage

get access to an cable clamp amp meter / watt meter from an electrician and read the current on starting the ac unit

if it exceeds 900 watts for 120 volts or 1150 for 240 volts , consider a bigger generator

for 240 volts drawing 4.8 amps the required wattage is 1152

for 120 volts the wattage is 576 watts

for 120 volts with a current of 900 watts , it equates to 7.5 amps

with 240 volts with a current of 1150 watts , it equates to 4.79 amps

so you can see the problem by not stating the supply voltage

get access to an cable clamp amp meter / watt meter from an electrician and read the current on starting the ac unit

if it exceeds 900 watts for 120 volts or 1150 for 240 volts , consider a bigger generator

Aug 31, 2017 | Fixya.com Heating & Cooling

If you can provide the Make

Jun 27, 2010 | AcerPower 1000 (AP1000EA380M) PC Desktop

No the LOWER the resistance the MORE Output or Draw, If say it is 100 Watts into 4 Ohms, if you used a 2 ohm speaker it would draw 200 watts.(only if capable of delivering that amount of power) The LOWER the Resistance, the HIGHER the Current Draw.
WATTS = AMPS x VOLTS. (This is not strictly true with amplifiers output though, as it is an AC Current, and a "Reactive" load, it has, Inductance, Capacitance and Resistance, all in one.and is worked out via a Formula)
see below.
http://en.wikipedia.org/wiki/Audio_power

http://www.hometheaterhifi.com/volume_1_1/v1n1spk.html

Now with Dual Voice Coils, Essentially, you can think of these speakers as TWO speakers in ONE enclosure. As such, They are basically TWO Speakers. Please see the website below it explains everything for you. http://www.diyaudioandvideo.com/FAQ/Wiring/

Everything is a function of Resistance and Ohms law.

Current (I) = Voltage (E) divided by Resistance(R).

Voltage = Current Multiplied by Resistance.

Resistance = Voltage divides by Current.

http://www.hometheaterhifi.com/volume_1_1/v1n1spk.html

Now with Dual Voice Coils, Essentially, you can think of these speakers as TWO speakers in ONE enclosure. As such, They are basically TWO Speakers. Please see the website below it explains everything for you. http://www.diyaudioandvideo.com/FAQ/Wiring/

Everything is a function of Resistance and Ohms law.

Current (I) = Voltage (E) divided by Resistance(R).

Voltage = Current Multiplied by Resistance.

Resistance = Voltage divides by Current.

Apr 17, 2010 | Car Audio & Video

there is no such thing as "ac/dc" current-its either one or the other--if its a 120 Volts then chances are it is AC already-at 60hz---the 15 Watts or any Watts is the amount of draw from the unit-not the power supplied to it- Watts= Volts(supplied)/Amps(being used) if you only want to supply 15 watts then reduce the amps (with a fuse) to .125amps

Mar 26, 2010 | Computers & Internet

They are probably blown. Here is instructions for testing a blown speaker.

Get yourself an electric meter, like a multi meter they sell at your local Radio Shack. Set it to read ohms (impedance). Make sure your speaker is off and put one meter lead on each of the two terminals. If your speaker is blown, it will read infinite impedance. If your speaker is good to go, it will read 1.0.

Get yourself an electric meter, like a multi meter they sell at your local Radio Shack. Set it to read ohms (impedance). Make sure your speaker is off and put one meter lead on each of the two terminals. If your speaker is blown, it will read infinite impedance. If your speaker is good to go, it will read 1.0.

Aug 13, 2009 | 2001 Honda Civic

Look at the rating sticker on the oven and use the AC power rating as your guide. The cooking power of the oven is not important, but the power it needs from the AC supply is. Voltage times current is the supply power needed (watts), and you want an oven that will draw less than the inverter's continuous rating, less any power being used by other appliances.

For instance, a microwave rated for 120 volts AC at 10 amps needs 1200 watts. Your inverter would be OK with this as long as other devices aren't using more than another 1000 watts. You can't use the inverter's peak rating. That's only for brief bursts, like when the microwave first kicks on and draws some extra current.

Any mid-power (600 watts or so) oven should be OK.

For instance, a microwave rated for 120 volts AC at 10 amps needs 1200 watts. Your inverter would be OK with this as long as other devices aren't using more than another 1000 watts. You can't use the inverter's peak rating. That's only for brief bursts, like when the microwave first kicks on and draws some extra current.

Any mid-power (600 watts or so) oven should be OK.

Apr 16, 2009 | Emerson MW8627W Microwave Oven

If you want to get more precise, figure out everything in terms of power (watts).

Basic electrical rule 1, 2 and 3:

voltage x current = power

or re-arranged:

current = power divided by voltage

or re-arranged:

voltage = power divided by current

For example, 12V X 2 amps = 24 watts.

or another example, 400 watts divided by 120 Volts = 3.33 amps

A 55W headlight that uses 12V would draw 55 /12 = 4.6 amps @ 12V

A 55 watt light bulb in a lamp at home would draw 55 / 120 = 0.46 amps @ 120V

As the previous post mentioned, inverters are not perfect when convertering 12V into 120V. If the converter consumes 1000W from the 12V battery, then a 90% effecient converter would generate 900W of 120V AC power best case. The other 100W is lost primarily as heat.

The other thing that gets tricky is that these ratings and the formula above are used for resistive loads, like light bulbs or hair dryers. Anything with a motor or transformer is considered an inductive load and can get much more tricky to calculate.

Consequently you need to give your self a safety margin when figuring out how big an inverter you need.

How does work in a practical sense?

Lets say you want an inverter for TV, DVD and Sat. Receiver. Look at the back of TV or in the manual. It should say how many watts it consumes. Lets say it is 400W. The DVD might be 100W and the Sat. receiver 50W - just as an example.

400 + 100 + 50 = 550 Watts. (just as an example)

You might think, well no problem, I'll use a 600 Watt inverter and have 50 watts left over. Depending on your inverter, that 600W might really be 600 x 90% effecient = 540 Watts of AC, less a 20% margin of error for the inductive transformers in the electronic of the TV, DVD and Sat. receiver 540 - 20% = 432 Watts.

Now you can see your 600 Watt inverter isn't big enough to do the job.

If we really need 550 watts of AC, add 10% to make up the effiency loss, then add a safety margin for inductive loads.

550 + 10% = 605 + 20% = 726 Watts.

Sounds more like an 800W inverter fits the job.

What does that mean in terms of wiring the 12V batteries to the inverter?

from the formula above:

current = power divided by voltage

In our example, we have an 800W inverter that runs on 12V

The current would thererfore be:

current = power divided by voltage

current = 800 watts divided by 12V

current = 66 amps.

That is important info because you can not use light gauge wire to carry 66 amps worth of 12V to the inverter nor could you use a 20A fuse to protect your inverter.

Now that's a lot of science for a guy who just wants to run a toaster on an inverter right?

800W / 120V = 6.66 amps

Using garryp's ratio 11:1, 6.66 x 11 = 73 amps.

That is a good ratio with a good safety margin.

This is all just MHO and should not taken as solid technical advise. In other words, don't blame me if you blow yourself up.

Basic electrical rule 1, 2 and 3:

voltage x current = power

or re-arranged:

current = power divided by voltage

or re-arranged:

voltage = power divided by current

For example, 12V X 2 amps = 24 watts.

or another example, 400 watts divided by 120 Volts = 3.33 amps

A 55W headlight that uses 12V would draw 55 /12 = 4.6 amps @ 12V

A 55 watt light bulb in a lamp at home would draw 55 / 120 = 0.46 amps @ 120V

As the previous post mentioned, inverters are not perfect when convertering 12V into 120V. If the converter consumes 1000W from the 12V battery, then a 90% effecient converter would generate 900W of 120V AC power best case. The other 100W is lost primarily as heat.

The other thing that gets tricky is that these ratings and the formula above are used for resistive loads, like light bulbs or hair dryers. Anything with a motor or transformer is considered an inductive load and can get much more tricky to calculate.

Consequently you need to give your self a safety margin when figuring out how big an inverter you need.

How does work in a practical sense?

Lets say you want an inverter for TV, DVD and Sat. Receiver. Look at the back of TV or in the manual. It should say how many watts it consumes. Lets say it is 400W. The DVD might be 100W and the Sat. receiver 50W - just as an example.

400 + 100 + 50 = 550 Watts. (just as an example)

You might think, well no problem, I'll use a 600 Watt inverter and have 50 watts left over. Depending on your inverter, that 600W might really be 600 x 90% effecient = 540 Watts of AC, less a 20% margin of error for the inductive transformers in the electronic of the TV, DVD and Sat. receiver 540 - 20% = 432 Watts.

Now you can see your 600 Watt inverter isn't big enough to do the job.

If we really need 550 watts of AC, add 10% to make up the effiency loss, then add a safety margin for inductive loads.

550 + 10% = 605 + 20% = 726 Watts.

Sounds more like an 800W inverter fits the job.

What does that mean in terms of wiring the 12V batteries to the inverter?

from the formula above:

current = power divided by voltage

In our example, we have an 800W inverter that runs on 12V

The current would thererfore be:

current = power divided by voltage

current = 800 watts divided by 12V

current = 66 amps.

That is important info because you can not use light gauge wire to carry 66 amps worth of 12V to the inverter nor could you use a 20A fuse to protect your inverter.

Now that's a lot of science for a guy who just wants to run a toaster on an inverter right?

800W / 120V = 6.66 amps

Using garryp's ratio 11:1, 6.66 x 11 = 73 amps.

That is a good ratio with a good safety margin.

This is all just MHO and should not taken as solid technical advise. In other words, don't blame me if you blow yourself up.

Nov 26, 2008 | Coleman 5640B807 Compact Refrigerator

May 26, 2018 | Magellan GPS

May 26, 2018 | GPS

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