Hi.

Bow and arrow.

There is a particular weight arrow that will fly further than a lighter or a heavier one for a given bow.

What is the matching named in the physics/mechanics world ? Is it COP coheficient of performance or am I peeing out of the can ?

Miguel

## Mechanical equivalent of impedance matching?

### Mechanical equivalent of impedance matching?

- Abolish the deciBel ! -

### Re: Mechanical equivalent of impedance matching?

There are examples of such matches, but I don't know of a common name which applies to all of them.

For example, when trying to get the best accelleration for a motor driven load in a servo system, a gear train can be used to make the reflected moment of inertia of the load match the moment of inertia of the motor armature.

The gearing on a bicycle is another example.

Also, when designing speaker enclosures, there may be an optimum volume for the enclosure for a given speaker.

<small>[ February 26, 2006, 10:04 PM: Message edited by: stephen ]</small>

For example, when trying to get the best accelleration for a motor driven load in a servo system, a gear train can be used to make the reflected moment of inertia of the load match the moment of inertia of the motor armature.

The gearing on a bicycle is another example.

Also, when designing speaker enclosures, there may be an optimum volume for the enclosure for a given speaker.

<small>[ February 26, 2006, 10:04 PM: Message edited by: stephen ]</small>

- Chris Smith
**Posts:**4325**Joined:**Tue Dec 04, 2001 1:01 am**Location:**Bieber Ca.

### Re: Mechanical equivalent of impedance matching?

Equilibrium.

I always call it Balancing my bow for the best performance of my choosing.

I studied my cross bow for months to get all of the best out of it.

I made darts that weighed just a few grams that went so fast that their kinetic energy transmitted into a mass that would blow a hole in a steel garbage can two inches wide, yet it lost all of that energy in an instant and didn’t even scratch the other side of the can.

I made Bolts that were weighted down and went through a brick wall and kept going.

But each bolt and dart had a draw back.

How far, and how fast, and what’s your purpose?

A bow merely accelerates a mass, the mass increases according to the bows efforts, and what you get is a result of these factors.

The bolt that goes through a brick will also stop short of 200 yards total flight.

The dart that couldn’t kill two sheets of steel, [only one] flies 400 yards or more depending on its feathers and lands like a feather.

Coefficient implies a set parameter to be...."Co" with "efficient"?

What is your "co" parameter, Maximum distance ONLY, Knock down power at what point in the flight, total speed?

If pure distance is your "CO" then the lighter one will always go the furthest, but a arrows feather design is everything as well.

Darts with feathers at the front and back fly the most, while some arrows have elongated feathers covering more than 50% of their total length to glide even further.

Again is your goal just gross yardage or usefulness?

<small>[ February 26, 2006, 09:09 PM: Message edited by: Chris Smith ]</small>

I always call it Balancing my bow for the best performance of my choosing.

I studied my cross bow for months to get all of the best out of it.

I made darts that weighed just a few grams that went so fast that their kinetic energy transmitted into a mass that would blow a hole in a steel garbage can two inches wide, yet it lost all of that energy in an instant and didn’t even scratch the other side of the can.

I made Bolts that were weighted down and went through a brick wall and kept going.

But each bolt and dart had a draw back.

How far, and how fast, and what’s your purpose?

A bow merely accelerates a mass, the mass increases according to the bows efforts, and what you get is a result of these factors.

The bolt that goes through a brick will also stop short of 200 yards total flight.

The dart that couldn’t kill two sheets of steel, [only one] flies 400 yards or more depending on its feathers and lands like a feather.

Coefficient implies a set parameter to be...."Co" with "efficient"?

What is your "co" parameter, Maximum distance ONLY, Knock down power at what point in the flight, total speed?

If pure distance is your "CO" then the lighter one will always go the furthest, but a arrows feather design is everything as well.

Darts with feathers at the front and back fly the most, while some arrows have elongated feathers covering more than 50% of their total length to glide even further.

Again is your goal just gross yardage or usefulness?

<small>[ February 26, 2006, 09:09 PM: Message edited by: Chris Smith ]</small>

### Re: Mechanical equivalent of impedance matching?

Simple lever.

### Re: Mechanical equivalent of impedance matching?

Your headline asks about impediance matching for mechanical systems and you cited bow and arrows.

Other possibilities would be spring strength and dashpot dampening (shock absorbers).

Most cars have the springs and shocks balanced such that the springs are not too stiff so the ride is not as hard nor are the springs too weak causing the car to bottom out. The shocks are there to dampen oscillations of the springs.

Another possible example would be a torque converter. This is a fluid filled device that couples mechanical energy from an engine to a transmission in a car. Think of two fans, one running and blowing the second. Only the fans are not in air but are in fluid. Too big a torque converter and energy is wasted in turning the torque converter. Too little and the engine will burn it up.

Other possibilities would be spring strength and dashpot dampening (shock absorbers).

Most cars have the springs and shocks balanced such that the springs are not too stiff so the ride is not as hard nor are the springs too weak causing the car to bottom out. The shocks are there to dampen oscillations of the springs.

Another possible example would be a torque converter. This is a fluid filled device that couples mechanical energy from an engine to a transmission in a car. Think of two fans, one running and blowing the second. Only the fans are not in air but are in fluid. Too big a torque converter and energy is wasted in turning the torque converter. Too little and the engine will burn it up.

No trees were harmed in the creation of this message. But billions of electrons, photons, and electromagnetic waves were terribly inconvenienced!

### Re: Mechanical equivalent of impedance matching?

Hi Jolly.

Yes, bow and arrow is the simplest analogy I could come up with to express impedance matching in mechanical form. Driver and driven. There is plenty of examples, I was looking for how it is named in physics/mechanics.

Chris: No particular purpose or usefulness, just trying to fill a knowledge void. It is COP cohefficient-of-performance, not CO. I have no experience at all with bows/arrows, just an easy example. If COP can be expressed in the ratio of different units, am not sure. It would be the flying distance divided by the force of the impulse.

A very light arrow (drinking straw?) will not fly further. Neither a heavy one (lead rod). There is one specific weight that will. Just as in electronics, the maximum power transfer match.

Miguel

Yes, bow and arrow is the simplest analogy I could come up with to express impedance matching in mechanical form. Driver and driven. There is plenty of examples, I was looking for how it is named in physics/mechanics.

Chris: No particular purpose or usefulness, just trying to fill a knowledge void. It is COP cohefficient-of-performance, not CO. I have no experience at all with bows/arrows, just an easy example. If COP can be expressed in the ratio of different units, am not sure. It would be the flying distance divided by the force of the impulse.

A very light arrow (drinking straw?) will not fly further. Neither a heavy one (lead rod). There is one specific weight that will. Just as in electronics, the maximum power transfer match.

Miguel

- Abolish the deciBel ! -

- Chris Smith
**Posts:**4325**Joined:**Tue Dec 04, 2001 1:01 am**Location:**Bieber Ca.

### Re: Mechanical equivalent of impedance matching?

EX,....

Co-Efficient is defined by at least two paramaters or more.

Co plus efficient, or efficient to what your trying to accomplish.

Distance, speed, streamline, drag......etc.

"Performance" is also defined by several individual paramaters as well.

Speed, distance, looks pretty,... etc.

Co-Efficient is defined by at least two paramaters or more.

Co plus efficient, or efficient to what your trying to accomplish.

Distance, speed, streamline, drag......etc.

"Performance" is also defined by several individual paramaters as well.

Speed, distance, looks pretty,... etc.

### Re: Mechanical equivalent of impedance matching?

Hi Externet.

I don't think there is a bow and arrow equivalent to impedance matching. With electrical impedance matching, maximum energy is transferred to the load when the load impedance is equal to the output impedance of the driver circuit. The driver's circuit's impedance absorbs an equal amount of energy.

With the arrow, you said you wanted to make the arrow fly farther. You have to be careful and note that an arrow that flies the farthest may not have any kill energy when it hits the target due to its low mass. NEGLECTING AIR DRAG, if you shoot a light arrow and then a heavy arrow straight up, the lighter one will stay aloft longer than the heavier one because it will be accelerated the fastest. Newton determined that the weight of falling objects does not apply to their rate of fall, so the lighter one will stay aloft longer merely because its speed made it fly higher.

I you want to determine the arrow which will leave the bow with the highest kinetic energy, but not necessarily fly longer, you have to use the formula E=(MV^2)/2 and maximize the arithmetical product of M which is the mass of the arrow and V squared which is the arrow's velocity squared. Increasing arrow mass M will obviously lower exit velocity V. IF M =0 then V will be high, but the product is zero. Same if V is 0 and M is high the product is zero.There is a balance here to get the maximum, just like impedance matching.

Before you start you'll be wanting to know how to determine V. It depends on the equations F=MA and V=AT. F is the bow string's draw force. Is that bow's draw force linear during all of it's travel? How long that travel is (how long is the draw?) because it will effect T, And does the moving mass of the bow effect the total value of M to any appreciable extent?

Yarrrgg! Someone help me! My mind is boggling, froth is coming out my ears, I haven't taken this high school crap in years!!! I should have paid more attention in calculus class!! What do I do with a non-linear draw force?

Then if you want to maximize the amount of energy of the arrow as it strikes its target, you have take into account energy loss due to air resistance and possibly increase the arrow's mass and/or decrease its air resistance. You'll probably find that arrows with different masses are more or less suitable for targets at different distances when optimizing maximum energy at the target.

This has most certainly been already done hundreds of years ago experimentally.

Bob

<small>[ February 28, 2006, 12:32 PM: Message edited by: Bob Scott ]</small>

I don't think there is a bow and arrow equivalent to impedance matching. With electrical impedance matching, maximum energy is transferred to the load when the load impedance is equal to the output impedance of the driver circuit. The driver's circuit's impedance absorbs an equal amount of energy.

With the arrow, you said you wanted to make the arrow fly farther. You have to be careful and note that an arrow that flies the farthest may not have any kill energy when it hits the target due to its low mass. NEGLECTING AIR DRAG, if you shoot a light arrow and then a heavy arrow straight up, the lighter one will stay aloft longer than the heavier one because it will be accelerated the fastest. Newton determined that the weight of falling objects does not apply to their rate of fall, so the lighter one will stay aloft longer merely because its speed made it fly higher.

I you want to determine the arrow which will leave the bow with the highest kinetic energy, but not necessarily fly longer, you have to use the formula E=(MV^2)/2 and maximize the arithmetical product of M which is the mass of the arrow and V squared which is the arrow's velocity squared. Increasing arrow mass M will obviously lower exit velocity V. IF M =0 then V will be high, but the product is zero. Same if V is 0 and M is high the product is zero.There is a balance here to get the maximum, just like impedance matching.

Before you start you'll be wanting to know how to determine V. It depends on the equations F=MA and V=AT. F is the bow string's draw force. Is that bow's draw force linear during all of it's travel? How long that travel is (how long is the draw?) because it will effect T, And does the moving mass of the bow effect the total value of M to any appreciable extent?

Yarrrgg! Someone help me! My mind is boggling, froth is coming out my ears, I haven't taken this high school crap in years!!! I should have paid more attention in calculus class!! What do I do with a non-linear draw force?

Then if you want to maximize the amount of energy of the arrow as it strikes its target, you have take into account energy loss due to air resistance and possibly increase the arrow's mass and/or decrease its air resistance. You'll probably find that arrows with different masses are more or less suitable for targets at different distances when optimizing maximum energy at the target.

This has most certainly been already done hundreds of years ago experimentally.

Bob

<small>[ February 28, 2006, 12:32 PM: Message edited by: Bob Scott ]</small>

**-=VA7KOR=- My solar system includes Pluto.**

### Re: Mechanical equivalent of impedance matching?

hi there

The word you are looking for is 'Inertial Matching'. This is a term from servotechnics describing the match between the Inertia of the load and the Inertia of the drive train. Inertial matching greatly influences the regulation capabilities of servosystems. Just do a Google lookup on the thing or drop me a line for more info

just my two cents

Da fripster

The word you are looking for is 'Inertial Matching'. This is a term from servotechnics describing the match between the Inertia of the load and the Inertia of the drive train. Inertial matching greatly influences the regulation capabilities of servosystems. Just do a Google lookup on the thing or drop me a line for more info

just my two cents

Da fripster

Once a WireHead, Always a WireHead

### Re: Mechanical equivalent of impedance matching?

How about "sweet spot"?

### Re: Mechanical equivalent of impedance matching?

chris Smith . . . has studied his bow for months . . . look at the fabulous amount of data and analysis he has produced . . .Less than 200 yds or more than 400 yds . fantastic . ..will he submit it for a PhD ? . . . A bolt (No weight given) . . . will go through a brick wall .. but would stop in 200 yds. . . .When a bow accelerates a mass . . THE MASS INCREASES according to the bow efforts. . .what did he do whehn he was . . . studying mechanical engineering (Not physics)

The real truth is that it is not possible to answer the original question without detailed knowledge of the Drag Co-efficients of the specific arrows. feathers do not directly increase the distance an arrow travels - they do so by creating drag, which keeps the arrows straight and the width of same prevents an arrow from Pitch and Yaw. Does not the very idea that a human (Chris included) could use a bow string (A device which might impart high velocity to a light arrow but which could hardly do much for a 1.0 lbf bolt) to impart enough energy to pass a bolt through a brick wall, suggest that, such a human could achieve the same by just throwing the brick at the wall ?

All of that aside, if you were to pull a bow string back through a distance of, lets' say, 1.0 ft and the average applied force would be say 10.0 lbf (I have no real idea how much it actually does take so I say 'average' because the force/energy applied to a bow string is by no means linear with distance - it takes much more force to draw a bowstring back from twelve inches to twelve and one quarter inches than it does to pull it from zero deflection to one quarter of an inch - simple mechanics - not in accordance with Hooke's Law) then the absolute energy you are going to apply to an arrow or missile is 10.0 ft.lbf (I notice that Chris now uses ft.lbf in place of, in spite of all his protests, lbf.ft) - Thus, if you had an arrow weighing (say) 4.0 oz (0.25 lbf) then the Kinetic Energy (KE) associated with its' flight would be 10.0 ft.lbf hence, since KE = w.v^2.2g its' velocity would be v = sqrt(2g.KE/w) = sqrt(2*32.174*10/0.25) = 50.734 ft/sec. Now!, if the arrow were fired level (They rarely are, for distance they are ususally angled upward to some degree) at say 5.0 ft above the ground then the arrow would commence to fall toward the ground at an acceleration of 32.174 ft/sec^2. The time it would thus take to reach the ground (Essentially unaffected by grad since the falling velocity would be low) would be equal to sqrt(2h/g) = sqrt(10/32.174) = 0.5575 secs. During the 0.5575 secs before the arrow encountered the ground it would have travelled (Assuming insignificant losses due to drag) 0.5575*50.734 = 28.3 ft. (I'm doing this right off the top of my head so I hope all of the figures are reasonably correct) Note that, having assumed insignificant losses to drag, up to the instant when the arrow stuck the ground it would still have possessed essentlially all of its' KE i.e. the ability to kill or maim. Clearly that doesn't fit the notion of 400 yds so let's try it for 4.0 grams per Chris i.e. v = sqrt(2*32.174*10/(4/453.59237)) = 270 ft/sec - In which case it would have flown 270*0.5575 = 150 ft. Of course, in practical circumstances, when trying to achieve distance the bow/arrow would clearly have been aimed upward. It is well known that, disregarding drag, to get such a propjectile to travel max distance, the accelerating device would be aimed upward at 45 dgrees above level. I am not going to do the math for how far that would cause the arrow to travel, or what angle would be needed for 400 yards - but I will if someone is interested.

If the (say) 1.0 lbf bolt were given the sdame treatment then its' ex bow velocity would be 25.37 ft/sec thus it would travel 14 ft before striking the ground. Come to think of it, if you took a quarter hammer (A bricklayers tool weighing 3.5 lbf) then you would have to raise it by 2.86 ft to acquire the 10.0 ft.lbf of KE. Could one break through a brick wall with such a stroke ? I think that, in order to break through such a wall with a 1.0 lnf bolt you might have to launch it with something like giant size skeet shooter catapult.

It is some what inappropriate to try to relate the subject form of energy or impedance matching, to that of electronics impedance matching. The object of electronics impedance is to obtain maximum energy transfer by matching source and load impedances, either having the opposite sign reactance to the other - according to the Maximum Power Theorem. However while such an arrangement transfers max possible power to the load - it provides equal energy dissipation in both source and load. If you applied the theorem to a 100 MW generator then, in order to apply 100 MW to the load, it would dissipate 100 MW in the generator (making it a 200 MW generator) such an arrangement, while working for 20 W audio, wouldn't be popular in power generation/distribution circles.

Chris clearly has read, or reads, a lot but seems to understand little of it. Would anyone, with as much to say as he, and a propensity for calling people fools, rear orifice talkers etc, not prove his points by math/logic ? . . . If he could ?

The real truth is that it is not possible to answer the original question without detailed knowledge of the Drag Co-efficients of the specific arrows. feathers do not directly increase the distance an arrow travels - they do so by creating drag, which keeps the arrows straight and the width of same prevents an arrow from Pitch and Yaw. Does not the very idea that a human (Chris included) could use a bow string (A device which might impart high velocity to a light arrow but which could hardly do much for a 1.0 lbf bolt) to impart enough energy to pass a bolt through a brick wall, suggest that, such a human could achieve the same by just throwing the brick at the wall ?

All of that aside, if you were to pull a bow string back through a distance of, lets' say, 1.0 ft and the average applied force would be say 10.0 lbf (I have no real idea how much it actually does take so I say 'average' because the force/energy applied to a bow string is by no means linear with distance - it takes much more force to draw a bowstring back from twelve inches to twelve and one quarter inches than it does to pull it from zero deflection to one quarter of an inch - simple mechanics - not in accordance with Hooke's Law) then the absolute energy you are going to apply to an arrow or missile is 10.0 ft.lbf (I notice that Chris now uses ft.lbf in place of, in spite of all his protests, lbf.ft) - Thus, if you had an arrow weighing (say) 4.0 oz (0.25 lbf) then the Kinetic Energy (KE) associated with its' flight would be 10.0 ft.lbf hence, since KE = w.v^2.2g its' velocity would be v = sqrt(2g.KE/w) = sqrt(2*32.174*10/0.25) = 50.734 ft/sec. Now!, if the arrow were fired level (They rarely are, for distance they are ususally angled upward to some degree) at say 5.0 ft above the ground then the arrow would commence to fall toward the ground at an acceleration of 32.174 ft/sec^2. The time it would thus take to reach the ground (Essentially unaffected by grad since the falling velocity would be low) would be equal to sqrt(2h/g) = sqrt(10/32.174) = 0.5575 secs. During the 0.5575 secs before the arrow encountered the ground it would have travelled (Assuming insignificant losses due to drag) 0.5575*50.734 = 28.3 ft. (I'm doing this right off the top of my head so I hope all of the figures are reasonably correct) Note that, having assumed insignificant losses to drag, up to the instant when the arrow stuck the ground it would still have possessed essentlially all of its' KE i.e. the ability to kill or maim. Clearly that doesn't fit the notion of 400 yds so let's try it for 4.0 grams per Chris i.e. v = sqrt(2*32.174*10/(4/453.59237)) = 270 ft/sec - In which case it would have flown 270*0.5575 = 150 ft. Of course, in practical circumstances, when trying to achieve distance the bow/arrow would clearly have been aimed upward. It is well known that, disregarding drag, to get such a propjectile to travel max distance, the accelerating device would be aimed upward at 45 dgrees above level. I am not going to do the math for how far that would cause the arrow to travel, or what angle would be needed for 400 yards - but I will if someone is interested.

If the (say) 1.0 lbf bolt were given the sdame treatment then its' ex bow velocity would be 25.37 ft/sec thus it would travel 14 ft before striking the ground. Come to think of it, if you took a quarter hammer (A bricklayers tool weighing 3.5 lbf) then you would have to raise it by 2.86 ft to acquire the 10.0 ft.lbf of KE. Could one break through a brick wall with such a stroke ? I think that, in order to break through such a wall with a 1.0 lnf bolt you might have to launch it with something like giant size skeet shooter catapult.

It is some what inappropriate to try to relate the subject form of energy or impedance matching, to that of electronics impedance matching. The object of electronics impedance is to obtain maximum energy transfer by matching source and load impedances, either having the opposite sign reactance to the other - according to the Maximum Power Theorem. However while such an arrangement transfers max possible power to the load - it provides equal energy dissipation in both source and load. If you applied the theorem to a 100 MW generator then, in order to apply 100 MW to the load, it would dissipate 100 MW in the generator (making it a 200 MW generator) such an arrangement, while working for 20 W audio, wouldn't be popular in power generation/distribution circles.

Chris clearly has read, or reads, a lot but seems to understand little of it. Would anyone, with as much to say as he, and a propensity for calling people fools, rear orifice talkers etc, not prove his points by math/logic ? . . . If he could ?

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