Why Does Doubling Speakers Double Sound Pressure with Less Power?

In summary: B SPL.So, the conclusion I reached is this: If you want to achieve + 6 dB SPL, you need to quadruple the power.
  • #1
PSuran
25
4
Hello,

I have a question that I've been struggling to find an answer to / understand. Probably because my understanding of physics is quite limited :)

Before I ask the question, these are the things I take as facts, so please correct me if something is wrong:

1. A speaker in half space (close to a wall), measures at 6 dB SPL higher than the same speaker in free field - because pressure doubles (True for the omnidirectional part of the spectrum). Neumann LINK

2. Two speakers, relatively close to each other, playing the same signal, measure at 6 dB SPL higher than each speaker individually - again, because pressure doubles. This is not true for uncorrelated signals. Sengpiel audio LINK

3. If I have one speaker, and I want to achieve +6 dB SPL, I need to quadruple the power (twice the power = +3 dB), this means using 4 times more electricity.

CONFUSION: But if we have two speakers side-by-side, as in one of the examples above, we get the same 6 dB boost by using only twice the electrical power.

To further clarify. The inverse-square-law explains that for a doubling of sound pressure we need 4 times the intensity. With intensity being "power per unit of area", and power being the rate of energy "usage" - this makes me conclude that having two speakers on simultaneously - uses twice the electrical power, but "creates" 4 times sound intensity, which, to me, seems paradoxical.

I'm sure I'm either misunderstanding something, or conflating some terms, but I hope that I've at least made clear what I'm misunderstanding. Any clarification would be much appreciated.
 
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  • #2
Overthinking it further :) :

Turning on two speakers is 6dB SPL more (assuming perfect summation) than each playing individually.

"Facts":

1. Driving two speakers requires two times electrical power.

2. 6 dBSPL increase is 4 times the intensity at the measurement device, which is 4 times the acoustic power ---> Maybe this is the wrong assumption, that power and intensity are linearly correlated. But i don't see why they wouldn't be. Even though intensity is "power per unit area", to have 4 times the power in a unit of area, we need 4 times the power at source. Even though not all the power reaches the "unit area", it is still a linearly proportional increase.

So, how can 4 times acoustic power be a result of 2 times electrical power?
 
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  • #3
PSuran said:
3. If I have one speaker, and I want to achieve +6 dB SPL, I need to quadruple the power (twice the power = +3 dB), this means using 4 times more electricity.

CONFUSION: But if we have two speakers side-by-side, as in one of the examples above, we get the same 6 dB boost by using only twice the electrical power. ...
I believe you are confusing power and pressure (SPL).

See: http://www.indiana.edu/~emusic/etext/acoustics/chapter1_amplitude4.shtml
A doubling of power equals an increase of +3dB.
a doubling of amplitude from one source to another equals an increase of +6 dB
 
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  • #4
Speakers work by moving air, if you have a rectified speaker setup then one speaker pushes and one speaker pulls (i.e. The negative and positive on the drivers are reversed), this moves proportionately more air without room compression for the same sine wave being reproduced.

This only works on a single channel and is solely dependent on the wiring within a cabinet for that single channel and will only work for an even number of speakers.
It is commonly used in guitar speaker cabinets.

Most quad boxes are wired this way and some guitar amps have a rectified multi speaker output as well.
I can't tell you the technical or algorithmic expressions, but I can tell you it is much louder and has a lot more punch.

I can't say I've ever seen it in stereo setups.
 
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  • #5
Hi NTL2009 and Idyit, thanks for the input, but that wasn't really what I was having issues with. I spent some time researching, and what I found was this:

Intro facts:
1. Double the energy = + 3dB
2. Double the amplitude = + 6 dB

Real-life experiment results:

Situation 1: One active speaker, measured @ 1m, @ 1 W of power, creates 90 dB SPL. To achieve + 6 dB, quadruple the power is necessary, meaning 4W. Measured to be true. Checks out.

Situation 2: Two times the same active speaker (meaning 2 speakers). Each speaker 1 W of power, SPL measured in the "sweet spot" (equal distance between both) is + 6 dB SPL. Measured to be true. Checks out.

My main "problem" was -> How come in situation 1 I am using 4 W total to achieve 96 dB, and in situation 2 I am using 2 W total to achieve the same 96 dBSPL.

In the meantime I found that "Situation 2" behaves like this.

+ 3 dB for double the power (remember, two active speakers)
+ 3 dB for double the cone size.

So the total is + 6dB.

This still doesn't account for the smaller energy bill in Situation 2, but I'm sure it has something to do with air impedance and what not, I just can't wrap my head around it yet :)
 
  • #6
What is the efficiency of a typical speaker? Are two more efficient than one? I'm not sure.

Cheers
 
  • #7
Just out of interest, how are you wiring the speakers and how do you achieve the same resistance for the total circuit ?

I suspect you are using more power in the two speaker system because the resistance is halved ?

How are you calculating the total power ?
 
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  • #8
Idyit said:
Just out of interest, how are you wiring the speakers and how do you achieve the same resistance for the total circuit ?

I suspect you are using more power in the two speaker system because the resistance is halved ?

How are you calculating the total power ?
I think these points are key.

From the OP's lnk, a different page, compares Electrical Parameters to Sound Parameters. This makes it clearer for someone like me with background in electronics:

http://www.sengpielaudio.com/calculator-ak-ohm.htm

Sound pressure p N/m² = Pa ≡ V or E voltage
Particle velocity v m/s ≡ I current
Acoustic impedance Z N·s/m³ ≡ R resistance
Sound intensity J or I
W/m² ≡ P power

When you double the voltage level into the same R, you get 4x the power (watts), as P = E2/R

See the above equivalencies, and with an acoustic impedance that remains constant (free field), doubling the sound pressure will give 4x the acoustic power.

It can be very difficult to make even roughly accurate acoustic measurements, due to phase differences, reflections, room interactions, and the interactions of two sources with supporting or cancelling wave points. It is normally done in a large an-echoic chamber, and it is still difficult.

But your issues may be simpler than that - I also am curious how you know you are inputting 1 W and 4 W. You really need to measure both Voltage and current, and take phase into account. Or simpler, you will be close dealing with ratios, and assuming speaker impedance remains fairly constant - and remember that if you double the Voltage to the speaker, the watts will be 4x (Z remained constant). For double the power, increase voltage by 1.414x.
 
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  • #9
Isn't this just an interference pattern? At some places you get 4 times the intensity, at other places you get nothing. If you integrate over a surface around both the speakers you should still get twice the power on average.
 
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  • #10
Idyit said:
Just out of interest, how are you wiring the speakers and how do you achieve the same resistance for the total circuit ?

I suspect you are using more power in the two speaker system because the resistance is halved ?

How are you calculating the total power ?

Disclaimer: I'm neither an engineer, nor a physicist, so just a note that the answer is probably really simple, or I am making some stupid mistake. Take that into account please :) And I really appreciate the effort.

To answer your above question. I am not actually measuring anything except SPL.

I "know" that I am using "double" electrical power in the two speaker setup because, well... I turn on the other speaker :) And at the same distance from both, each individually measure an SPL value 6 dB lower than both together. Is this the dumb assumption? :)

That's what meant when I wrote that I measured the results. But to be honest, people smarter then me have told me that if I have a single speaker and I boost the volume until I measure a 6 dB increase in SPL - electrical power consumption has increased four times.
 
  • #11
And this:

"In the meantime I found that "Situation 2" behaves like this.

+ 3 dB for double the power (remember, two active speakers)
+ 3 dB for double the cone size.

So the total is + 6dB."

This is the info I found in literature... Not by experimentation. I can link the text here if it helps...
 
  • #12
Yup I'd agree with that ;-)
 
  • #13
Idyit said:
Yup I'd agree with that ;-)

With what exactly? :)
 
  • #14
+3db from the halved impedance (i.e twice the current)
+3db from twice the cone size (i.e twice the amount of air movement)
 
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  • #15
Idyit said:
+3db from the halved impedance (i.e twice the current)
+3db from twice the cone size (i.e twice the amount of air movement)

Yes, but the part I'm struggling to understand is how is it possible that I have half the electricity bill with the 2 speaker setup, vs. the one speaker setup at same SPL.

Because with the 2 speaker setup I double the electrical power (turn on two speakers), but quadruple the acoustic power (a measured 6 dB increase = 4 times power). Wouldn't this mean more energy out than it came in? :)
 
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  • #16
You electricity bill will double if you are running the same volume on the amp.

The mysteries of more air movement have been around along time, I guess the only answer is you are converting the power much more efficiently using two speakers.
 
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  • #17
Yeah, thinking off it... I think the answer to the following sub-question would put me out of my misery :)

With power being constant, simply doubling the speaker size (cone area), increases the output for 3 dB. But doesn't the increased size come with downsides? Heavier, harder to move, more air resistance... Which would require more power to compensate?
 
  • #18
I doubt air resistance would make any difference, The magnets in the drivers might make a small difference.

All I can tell you is more speakers equals a much bugger sound, working with a Band for 20years, If I had two quad boxes plugged into the same amp I could run the amp at a much lower volume and get the same sound on stage, I can't tell you the fluid dynamics / physics equations for it, but I know it from standing in front of it :-)
 
  • #19
PSuran said:
Yeah, thinking off it... I think the answer to the following sub-question would put me out of my misery :)

With power being constant, simply doubling the speaker size (cone area), increases the output for 3 dB. But doesn't the increased size come with downsides? Heavier, harder to move, more air resistance... Which would require more power to compensate?
But power wouldn't be constant. If you double the cone size, and keep the motion distance the same, you are moving more air, and that takes more power.

If you double the cone size, and then reduce the motion distance to get to the same SPL, you would have brought the power back to the original level.

All that is an over-simplification, as speakers/drivers are optimized to work in a certain range, and you just can't change things like tha , they need to be designed for it. But yes, connecting two identical speakers in parallel, with an amp capably of driving both loads equally well, you should have double the watts in (same voltage 2x current so 2x watts), and double the sound power out. Again, I think you are confusing pressure levels (~ voltage) with power levels (~ watts). Pressure only needs to increase 1.414x for a doubling of power. Assuming constant acoustic impedance.
 
  • #20
NTL2009 said:
Again, I think you are confusing pressure levels (~ voltage) with power levels (~ watts). Pressure only needs to increase 1.414x for a doubling of power. Assuming constant acoustic impedance.

Well, I'm not sure I'm confused about what you say. I understand that + 6dB is twice the voltage (pressure), but 4 times the power.

NTL2009 said:
But power wouldn't be constant. If you double the cone size, and keep the motion distance the same, you are moving more air, and that takes more power.

If you double the cone size, and then reduce the motion distance to get to the same SPL, you would have brought the power back to the original level.

Well, exactly! But, to account for the + 6 dB when we add "another active speaker", the sources say that + 3 dB is for double the electrical power used, and the additional + 3 dB is for double the cone size (6 dB total). And I don't understand that logic for the reason you stated (if I understand correctly) -> Increasing the cone size uses more energy, requires more power to "move", if dB gain is required.

Meaning, all other parameters remaining constant, I don't see how simply increasing cone size gives me "free 3 dB". I'm sure it isn't free, I just don't understand where it comes from :)

A qoute from the web as an example: "Recently upgraded (more low end) and downsized my subs, going from a horn loaded design to ported cabs. Reduced size by 25%, decreased weight by 5%, increased LF output by 6dB, using 3dB more power, which happened to be available from the amps I was using.

Re-built 4 cabinets with 2 Lab 12s in each. As predicted, doubling the cones adds 3dB sensitivity, and was verified by tons of real world testing slogging cabinets across the snow, mud and coal dust. "
 
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  • #21
One thing I'll add is an outdoor setting may change your results, In all the outdoor gigs I've done the levels dropped radically for the same power amps.
I think NTL2009 is on the right track, twice the cone size in an enclosed space is twice the air movement/pressure being generated, outdoors the equation will be very different.
 
  • #22
Idyit said:
One thing I'll add is an outdoor setting may change your results, In all the outdoor gigs I've done the levels dropped radically for the same power amps.
I think NTL2009 is on the right track, twice the cone size in an enclosed space is twice the air movement/pressure being generated, outdoors the equation will be very different.

Exactly, because having a wall behind the speakers adds 6 dB - because it radiates into half space. Placing a sub on the floor in the corner (quarter space) can add 12 dB or more... Having other reflections also adds "volume". And considering 10 dB is subjectively evaluated as doubling the loudness, it's clear why you need more power outdoors.

But, adding a second active speaker adds 6 dB even in free field. That's what literature says.
 
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  • #24
PSuran said:
I think I found the answer, but the problem is that I don't understand it :) :)

http://www.linkwitzlab.com/faq.htm#Q21
It's confusing to me too, but Linkwitz is an expert in the field (I'm familiar with his work) so I'll accept that explanation that it is the doubling of surface area that is increasing the efficiency of electrical to acoustic conversion.

So if we accept that, it's clear that in the case with a single speaker, you need to double the electrical power into double the acoustic power out, because you didn't change the surface area. Then the comparison works, right?

Though I'm still not grasping how two individual, identical speakers which put out Y acoustic power each with X electrical power in can somehow create 4x the acoustic power. His formulas say so, but I don't get it. How can they act differently together versus separate (unless they are being coupled somehow, but I see no mention of that?).
 
  • #25
NTL2009 said:
Though I'm still not grasping how two individual, identical speakers which put out Y acoustic power each with X electrical power in can somehow create 4x the acoustic power.

Exactly, that's the thing that's bothering me.

If we look at it from another angle...

1. One, perfectly omnidirectional speaker in free field radiates 100 dB in all directions (point source - spherically) at a low enough frequency.
2. If we now add one dimension, mount the speaker into a wall, it now radiates all the energy into half the space (Let's assume no deconstructive interference - which is true at a low enough frequency).
3. This now means that we have doubled the radiating power in front of the wall, which would mean + 3dB.
4. But we get + 6 dB! That's measured fact, and all sources say it behaves like so.

And most simply say, "yes because we are adding pressures and not power"... But if you check my above example, you can understand why this explanation is a bit handwavy. We are always talking about energy in some form, and in this case, it is DOUBLE the energy, not FOUR TIMES, which would be necessary for doubling the pressure.

So I'm sure there is a reason for this increased efficiency, I just don't get it yet :) And this reason is not "because pressure is doubled", yes, pressure is doubled because power is quadrupled.
 
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  • #26
Could it be this simple...

When I turn one speaker on, it creates sound, but increases air impedance while doing so. When I turn on another speaker with the same power, it "works into" this increased impedance. This increase in impedance when two speakers are on is what accounts for the increased efficiency?
 
  • #27
I was thinking about it today from a different angle - the logarithmic nature of power efficiency.Everything we've been saying implies that:

- Using two speakers with same total amount of power (2 x 50W) is more efficient than one speaker with same power applied (1 x 100W).
- Speakers are "driven by voltage". Not sure I can visualize this, but hey. I believe it :)

The formula which relates power and current is: U squared = P/R

Let's test what happens in the example with our two speakers.

If:

Two speakers, 500W each:

R = 10 (let's say) ohm
P = 500 W
U = 70.7 Volts

So we have two speakers driven by 70.7 V each.

One speaker driven with the same total amount of power (1000 W):

R = 10 (let's say) ohm
P = 100 W
U = 100 Volts

So the single speaker is driven by 100 V.

If we compare 141.4V and 100V using the dB formula (dB = 20 log V1/V2), we get:

dB = 20 log 141.4 / 100, which is...

EXACTLY 3 dB


I'm just not sure how to visualize all this :)

P.S.:

The analogy which got me thinking like this is:

Let's say you have to transfer an X number of passengers (more than fits in one car) from A to B in the smallest amount of total time, with most efficient fuel usage.

Is it better to use one car driving faster, or two cars driving slower?

Considering (let's say) 50 horsepower is used to drive 100kmph, and 100 horsepower is used to drive 120kmph, it is much more efficient to just drive slower, and use two cars! (If you have two "free" cars at your disposal, and fuel is the only concern)
The exact numbers used in the analogy are wrong, but the point is clear I guess. We used less total fuel AND transferred all the passengers sooner.
 
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  • #28
Where the two pressure waves intersect you are adding the two pressures together (bright spots), but you are also subtracting when they are opposite giving no pressure (dead spots). This will balance out so there is conservation of energy. You will have double pressure (4 times power since it is proportional to the square of the pressure) in some spots and zero pressure in others. The average power of the system stays the same as your input power. The analogy should be closer to waves on water intersecting. Sound engineers need to be careful of speaker placement because it will cause bright spots and dead spots because of this.
 
  • #29
Kinn said:
Where the two pressure waves intersect you are adding the two pressures together (bright spots), but you are also subtracting when they are opposite giving no pressure (dead spots). This will balance out so there is conservation of energy. You will have double pressure (4 times power since it is proportional to the square of the pressure) in some spots and zero pressure in others. The average power of the system stays the same as your input power. The analogy should be closer to waves on water intersecting. Sound engineers need to be careful of speaker placement because it will cause bright spots and dead spots because of this.

Thanks for the input. I fail to see how this can be the case. If we take into account the fact that we can never get more than +6 dB if two speakers are combinted, consider these scenarios:

Situation 1: Two subs one next to the other. Play a 30 Hz sine wave. You pretty much have +6 dB in all directions, everywhere (low enough wavelength, sources close enough).
Situation 2: Now place them 2.8m apart. You have +6 dB at equal distance from both, but huge dips at other locations.

If, as you say, the balance of dead spots to brighter spots was the only important thing, we would need spots "brighter" than 6 dB in situation 2. That just doesn't happen.
 
  • #30
Good point, let me see if I can figure it out in the math.
 
  • #31
PSuran said:
Thanks for the input. I fail to see how this can be the case. If we take into account the fact that we can never get more than +6 dB if two speakers are combinted, consider these scenarios:

Situation 1: Two subs one next to the other. Play a 30 Hz sine wave. You pretty much have +6 dB in all directions, everywhere (low enough wavelength, sources close enough).
Situation 2: Now place them 2.8m apart. You have +6 dB at equal distance from both, but huge dips at other locations.

If, as you say, the balance of dead spots to brighter spots was the only important thing, we would need spots "brighter" than 6 dB in situation 2. That just doesn't happen.
1)The interference pattern is peaked, not sinusoidal, so the area above 0dB is less than half the total.
2)If you place two speakers close together, they are pushing the same piece of air. So each one has an easier time and absorbs less power from the generator. For this reason, no increase in sound field is obtained. Another way of looking at it is to say that each speaker is receiving power from the other, and when you measure the speaker impedance it is altered so that each speaker takes less power.
 
  • #32
tech99 said:
1)The interference pattern is peaked, not sinusoidal, so the area above 0dB is less than half the total.

Sorry, I don't understand this :( Please clarify.

tech99 said:
2)If you place two speakers close together, they are pushing the same piece of air. So each one has an easier time and absorbs less power from the generator. For this reason, no increase in sound field is obtained. Another way of looking at it is to say that each speaker is receiving power from the other, and when you measure the speaker impedance it is altered so that each speaker takes less power.

This makes perfect sense, and was my initial understanding. The only reason why I'm having issues with this explanation is the following:

I have measured (in live sound and studio situations) that the ONLY REQUIREMENT for the + 6 dB SPL increase is that the measurement device (SPL meter) is in the triangle apex (same distance, sweet spot) between the speakers. It doesn't matter how far away the speakers are one from the other.

So I would be inclined to conclude that what you wrote above is true, but that this impedance change is inherent to the point of summation (measurement), regardless of the source. What do you think?
 
  • #33
The in-phase addition will always occur along the centre line, and in other places depending on geometry.
The impedance change will occur only when the speakers are a small fraction of a wavelength apart.
 
  • #34
tech99 said:
The in-phase addition will always occur along the centre line, and in other places depending on geometry.
The impedance change will occur only when the speakers are a small fraction of a wavelength apart.

OK then, try visualizing this:

Free field -> We have two speakers 2m apart. The SPL meter is at the same distance from both, at the triangle apex. If I understand your point, for a low enough frequency, we should be measuring + 6 dB (double power + impedance change). Also, if we just scale this setup up -> increase the distance between the speakers (to, let's say, 15 meters), and keep the SPL meter in the triangle apex - we will lose the "impedance" effect?
 
  • #35
Having read most of these posts I have to conclude that sound must be one of the worst possible phenomena to discuss to help you get an understanding of Interference. The programme material covers a range of around ten octaves. The fields at the location of the listener are a combination of very near field, intermediate field and far field. The simple two-source interference pattern of Young's Slits just doesn't apply.
The only thing one can rely on is that two identical sources of Power will produce 3dB more total power than a single source and up to 6dB more in some places - as long as the mutual effects between the two sources are negligible. The power cannot be greater and it all 'has to go somewhere'. There is no Magic at work.

One of the best ways to learn about the behaviour of Waves is to start with the most ideal situation and that is almost certainly what you get with short Radio Frequency Dipoles, at a single frequency, with fields sampled with a short dipole probe. Dipoles are not ideal (omnidirectional and polarised) but neither are loudspeakers or microphones (or ears) ideal. If you get familiar with this stuff then you are less likely to make the sort of gaffs that are so likely when talking about sound phenomena.
 
<h2>1. Why does doubling the number of speakers result in a higher sound pressure?</h2><p>When sound waves are produced by a speaker, they spread out in all directions. This means that the sound energy is distributed over a larger area, resulting in a lower sound pressure. However, when you double the number of speakers, the sound waves from each speaker overlap and combine, resulting in a higher sound pressure at a specific location.</p><h2>2. How does doubling speakers result in less power being used?</h2><p>When you double the number of speakers, each speaker does not need to work as hard to produce the same sound level. This is because the sound waves from each speaker combine and reinforce each other, resulting in a more efficient use of power. Therefore, less power is needed to achieve the same sound pressure as a single speaker.</p><h2>3. Is there a limit to how many speakers can be doubled before there is no change in sound pressure?</h2><p>Yes, there is a limit to how many speakers can be doubled before there is no significant change in sound pressure. This is because at a certain point, the sound waves from the speakers will start to cancel each other out, resulting in a decrease in sound pressure. The exact number of speakers that can be doubled before this happens will depend on various factors such as the size and layout of the room.</p><h2>4. Does doubling speakers always result in a better sound quality?</h2><p>No, doubling speakers does not always result in a better sound quality. While it may increase the sound pressure, it can also lead to distortion and uneven sound distribution if the speakers are not properly placed and calibrated. Additionally, the quality of the speakers themselves also plays a significant role in the overall sound quality.</p><h2>5. Can doubling speakers be used as a substitute for increasing the volume?</h2><p>Yes, doubling speakers can be used as a substitute for increasing the volume to some extent. However, it is important to note that doubling speakers does not necessarily mean doubling the volume. Additionally, as mentioned before, doubling speakers can also lead to distortion and uneven sound distribution, so it may not always be the most effective solution for increasing volume.</p>

Related to Why Does Doubling Speakers Double Sound Pressure with Less Power?

1. Why does doubling the number of speakers result in a higher sound pressure?

When sound waves are produced by a speaker, they spread out in all directions. This means that the sound energy is distributed over a larger area, resulting in a lower sound pressure. However, when you double the number of speakers, the sound waves from each speaker overlap and combine, resulting in a higher sound pressure at a specific location.

2. How does doubling speakers result in less power being used?

When you double the number of speakers, each speaker does not need to work as hard to produce the same sound level. This is because the sound waves from each speaker combine and reinforce each other, resulting in a more efficient use of power. Therefore, less power is needed to achieve the same sound pressure as a single speaker.

3. Is there a limit to how many speakers can be doubled before there is no change in sound pressure?

Yes, there is a limit to how many speakers can be doubled before there is no significant change in sound pressure. This is because at a certain point, the sound waves from the speakers will start to cancel each other out, resulting in a decrease in sound pressure. The exact number of speakers that can be doubled before this happens will depend on various factors such as the size and layout of the room.

4. Does doubling speakers always result in a better sound quality?

No, doubling speakers does not always result in a better sound quality. While it may increase the sound pressure, it can also lead to distortion and uneven sound distribution if the speakers are not properly placed and calibrated. Additionally, the quality of the speakers themselves also plays a significant role in the overall sound quality.

5. Can doubling speakers be used as a substitute for increasing the volume?

Yes, doubling speakers can be used as a substitute for increasing the volume to some extent. However, it is important to note that doubling speakers does not necessarily mean doubling the volume. Additionally, as mentioned before, doubling speakers can also lead to distortion and uneven sound distribution, so it may not always be the most effective solution for increasing volume.

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