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My thought was a relief cut was needed. Not just to set sear engagement length re. hook height but maybe to cam out the sear as the sear's escape edge passes under the hammer hook tips. I also thought today's hammers might need that relief to ensure the sear wasn't recaptured by the half cock notch as the hammer dropped. This issue could be more pronounced for a bullseye shooter over an action shooter as the bullseye shooter has a much slower trigger pull.
The overtravel screw must be set to prevent the sear nose from catching the half cock notch. Not too hard to do.

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I don't measure the torque, I calculate it from the measured weight.
Yes, my mistake. We are measuring force.

Excellent... You saw the same problems with the Jig as I and fixed them. The Ed Brown Jig is similar with similar problems. You saw the supplied shim as wrong (it needs to be 0.035" for Ed Brown not 0.020") and something different is needed to ensure the stone is parallel to the long top flat of the jig. The tensioning spring is a clever adder. Screws are terrible positioning devices unless loaded as there is slop in threads (both vertical and horizontal). The precisely machined face you are using for a reference to exactly locate the escape edge above sear pin center is spot on. I'm doing the exact same thing.

With those fixes I think the jig will be very accurate and highly repeatable.
Thanks, it does work really well. You should also get accurate results with your modified jig.

I also have a low cost 40x "pen" microscope with a built in .001" graticule. The sear fixture sits in a machine vise, and the microscope sits in a small microscope stand sitting nearby. As I cut the sear face, I can periodically just slide the vise and fixture under the microscope to inspect the progress.

So here is a funny. The Ed Brown Jig states the set screw should a specific distance, +/- 0.001. One finds in solids modeling, with an ordnance sear, that specific distance places the escape edge directly above the sear pin sear. Great. But nobody uses ordnance sears with a sharp point at the escape edge. Nearly everybody cut the 45 deg relief. So the sear angle that is then created by the jig is greater then 90 deg. Maybe that is the intent, or maybe they don't understand.
That is an example of what I mentioned about 90 degree sears being referenced to a 90 degree vertex .030" behind the front sear edge. That seems to be the only convention for specifying sear face angle.

My "87" and "88" sears are as they might be measured from that .030" 90 degree vertex. That was just a way of distinguishing them from the "90" sears.

My "88" sear is in fact 90 degrees as measured from the escape edge.

Vectors can always be added together. I need to see diagrams of your analysis before I can comment further. You may have shown them in the next post you made.
Ok. I suspected they might be, but wasn't sure with mechanical forces.

When you ask if the static and sliding coefficients track I assume you mean do they have a fixed proportionality between them. The answer is generally yes, but we may not be able to discover that constant. Also I hate to think the sliding coefficient might be velocity dependent. That is a different value for the bullseye shooter from the action shooter. Specialized instruments is likely beyond our means. In the end we may have to settle for a bounded solution rather then a fairly specific number.
I tried a crude experiment. I pulled my machine vise across the benchtop with a pull gauge. The benchtop has a fairly smooth laminated surface. There was a definite initial pull force due to static friction that dropped once the vise began to move. Beyond that, I could not detect any significant variation in the sliding friction force at various pull speeds.

There was some eratic readings at very slow speeds and due to pull spring reactance. So there will be a minimum velocity dependant on the pull gauge spring rate. But it appears that the sliding friction force is independant of velocity.

Within this thread someone mentioned a trigger measurement system which had a loadcell, likely a ball screw, and stepper motor to move the trigger in a controlled matter. The system continuously recorded force and displacement. I have designed and built such devices in the past. This one is German made, which generally means it is high quality. However, for a German product there is a surprising lack of any performance information. The one spec they provide states they measure (sense?) force to 1 part in a million (re full scale). Thus their claimed signal to noise ratio is 120 dB. This is basically impossible to achieve in a mechanical system.
Signal averaging is one way to eliminate noise. This method is apparent in gauges that have slow update rates. For a triigger pull measurement, I am guessing that you would need to complete the pull in less than a second. Possibly a lot less. So you would need a sample aquisition rate of better that 200 samples/sec for a .020" sear face. Still plenty of time to over sample and average with modern circuitry. May require a custom built sytem though. The details would be in the strain gauge bridge amplifier and the signal processing.

Not impossible though. 30 years ago, I worked on systems that measured femto coulombs of charge. And just one shot at the measurement.

Loadcells can be stiff, but cheap ones aren't as they cost more. As you point out we need a way to measure both force AND displacement simultaneously. The displacement part is tricky. To ensure accurate displacement measurement one doesn't want any of the systems load force to pass through the displacement sensor. That is load force must be shunted around the displacement sensor. I don't think the German design does this. They likely said we have a ball screw, it has so many turns/inch. We have a stepper motor, we can easily count teeth. We now have a displacement sensor. Yeah.... but it won't accurately resolve 0.0001" which is what we need.
As a drive, I think stepper motors should work if the step size is fine enough, and there is enough mechanical filtering to provide a smooth pull. Otherwise, the sear will pause at the points of stepper motor increments, and reach new points of static friction.

As you noted, there will be a discrepency between the actuator position and the sear position. This can be compensated for if you can determine the load cell spring rate. It should be linear. Your values of sliding friction would be slightly displaced along the x axis (in a graph) and by an easily calculated amount.

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Discussion Starter · #283 ·
Yes, that is the basic principle.

My method of analysis for the cam force uses simple mechanics, and goes straight to the end results. A hammer hook, under a certain tension, moves by a certain amount, while a sear moves by a certain amount. That's everything we need to know. Everything involved is incorporated into this result.

This illustration may help:

View attachment 611756

The calculations and the terms would be as follows:

F_hook is the hook force as derived from hammer torque measurements.

alpha is the hook displacement angle relative to the sear axis location.

dx_sear is .0001". Results are more accurate as this term becomes smaller. But the size is limited by the precision of the math functions.

x1_sear is the point along the sear face that is directly beneath the hammer hook tip. It is varied in .001" increments from the front edge of the sear to the escape edge. This point is relative to the 90 degree vertex point.

x2_sear = x1_sear - dx_sear

y1_sear = square root(x1_sear² + .4055²)

y2_sear = square root(x2_sear² + .4055²)

dy_sear = y1_sear - y2_sear

dy_hook = dy_sear / cosine(alpha)

F_cam = F_hook * cosine(alpha) * dy_hook / dx_sear

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I believe your analysis is as I describe below. I may not be interpreting your thoughts correctly.

You have an x-y coordinate system which is fixed to the sear. This coordinate system rotates relative to fixed space. The center of the coordinate system is the sear's center. The x axis runs along the face of the sear where +x is directed to the left and +y is upward.

x_sear is the current location of the hammer hook tip. You decrement x_sear by a fixed amount equal to dx_sear till x_sear=0 at the escape edge. The escape edge is at the 90 deg vertex of the sear. dy_sear is equal to the difference between y_sear_1 and y_sear_2.

I see some things which may need further investigation.

You are calculating y_sear assuming the sear has a 90 deg sear angle. (ie y_sear is the long leg of the right triangle). So in its present form the analysis is limited to 90 deg sears.

You are assuming the hypotenuse of your right triangle is a constant 0.4055 for all calculations of y_sear. This is only true when the hammer tip is at x_sear=0.030. The hypotenuse varies with the value for x_sear getting smaller with each smaller value for x_sear. Separately, for a 90 deg sear, y_sear will be constant and dy_sear=0

You are calculating y_sear=sqrt(x_sear^2+ Hyp^2), where Hyp is a constant. So dy_sear/ dx_sear= 1/2(x_sear^2+Hyp^2)^(-1/2)(2x_sear)= x_sear/(x_sear^2+Hyp^2)(1/2).

For dy_sear/ dx_sear= x_sear/(x_sear^2+Hyp^2)(1/2) we looking at the two terms under the sqrt. As x_sear^2<<<Hyp^2 this causes dy_sear= x_sear/Hyp. As x_sear approaches 0 at the escape edge dy_sear also approaches 0.

The cam force is F_cam= cos(alpha) F_hook dy_hook/dx_sear where dy_hook=dy_sear/cos(alpha).

Upon substitution, F_cam reduces to F_hook (dy_sear/dx_sear) with no dependence on the angle alpha.

As x_sear approaches 0 (the escape edge) the value for dy_sear also approaches zero (dx_sear is the non-zero incrementor), this causes F_cam to also approach zero. Yet we know F_cam doesn't go to zero at the escape edge.

Again I may not be correctly understanding your analysis.
 

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I find it so telling that the OP has stated earlier (post #168) that the only thing keeping the hammer cocked without the sear spring is friction, when the geometry shows otherwise and when this relationship is shown he simply ignores it. Or that the "relief cut" is used to keep the half cock from catching the sear,(Post #281) an obvious novice to proud of his math skills to see reality. Or when pulling his trigger very slowly (Post #169) there is a hesitation, that he believes ultra grease is the cure.

It is a shame when this happens, as his blindness and refusal to listen to others for the purpose we come here for. Understanding, without an open mind there can't be any. Sad, as this forum is a treasure trove of members with very good information, and are willing to share with anyone willing to listen. Resistance to listen is the antithesis of understanding.

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Discussion Starter · #285 ·
I also have a low cost 40x "pen" microscope with a built in .001" graticule. The sear fixture sits in a machine vise, and the microscope sits in a small microscope stand sitting nearby. As I cut the sear face, I can periodically just slide the vise and fixture under the microscope to inspect the progress.
That 40x pen scope is a great idea. Is that the one from Brownells? I have seen it many times. I don't know why I didn't think of that. I'm getting slow.

That is an example of what I mentioned about 90 degree sears being referenced to a 90 degree vertex .030" behind the front sear edge. That seems to be the only convention for specifying sear face angle.

My "87" and "88" sears are as they might be measured from that .030" 90 degree vertex. That was just a way of distinguishing them from the "90" sears.

My "88" sear is in fact 90 degrees as measured from the escape edge.
So now the million dollar questions...

Have people been getting poor trigger jobs because the instructions that come with the various jigs assume sears have sharp points at the edge edge??

Also all the jigs I have seen bank the sear against a feature on the jig (the set screw). But what about the feature on the sear that banks against the jig. All these sear features only touch air. Why should a sear manufacturer hold one of these feature to +/- 0.001 or even +/- 0.005" . There is no inspection of these parts to see if ALL dimensions are within spec. If a manufacturer missed a dimension no one would know as the feature only touches air and thus the sear works fine. These sear features have a much shorter distance to sear center then the sear's flat. Thus an error on the sear feature magnifies the error in location of the 90 deg vertex.

I tried a crude experiment. I pulled my machine vise across the benchtop with a pull gauge. The benchtop has a fairly smooth laminated surface. There was a definite initial pull force due to static friction that dropped once the vise began to move. Beyond that, I could not detect any significant variation in the sliding friction force at various pull speeds.

There was some eratic readings at very slow speeds and due to pull spring reactance. So there will be a minimum velocity dependant on the pull gauge spring rate. But it appears that the sliding friction force is independant of velocity.
The erratic reading was likely slip-stick induced by the low stiffness of the spring in your pull gage. One pulls on the gage and the spring elongates. The elongation of the spring is such that its force matches then exceeds static friction. The object breaks free and now a lower dynamic friction occurs. At the instant the spring is still applying the force needed to exceed static friction. The force imbalance on the object has the object to accelerate. The object temporarily exceeds the velocity you are pulling the gage. The elongation of the spring drops, the spring's force drops, spring force now is below that for dynamic friction, the object stops moving, the cycle repeats.

We may not have the tools to resolve a speed dependence to dynamic friction.

Signal averaging is one way to eliminate noise. This method is apparent in gauges that have slow update rates. For a triigger pull measurement, I am guessing that you would need to complete the pull in less than a second. Possibly a lot less. So you would need a sample aquisition rate of better that 200 samples/sec for a .020" sear face. Still plenty of time to over sample and average with modern circuitry. May require a custom built sytem though. The details would be in the strain gauge bridge amplifier and the signal processing.

Not impossible though. 30 years ago, I worked on systems that measured femto coulombs of charge. And just one shot at the measurement.
Signal averaging will definitely reduce noise. However it relies on a repetitive signal to average out the random noise. Thus your one shot measurement was likely difficult to get as it was non-repetitive.

We could over-sample our one shot trigger pull but care is needed. The trigger is moving during the sampling and if not careful we could average out useful information. Also noise could be in-band and thus indistinguishable from actual signal. Sampling here acts as a low pass filter. We don't want to set the filter to low and loose signal. These are the issues that drive experimentalists mad.

If the noise is purely electronic I have found battery powering the sensors with good shielding and single point grounds cut it down a lot without resorting to electronic (or digital) filters or over-sampling.


As a drive, I think stepper motors should work if the step size is fine enough, and there is enough mechanical filtering to provide a smooth pull. Otherwise, the sear will pause at the points of stepper motor increments, and reach new points of static friction.
It isn't really an issue of step size as the motor can be geared down. Also the pulses sent to the stepper can have the motor turn fast enough that sear never pauses.

The problem with stepper motors is it is like dead reckoning navigation. It kinda goes like this. I send an electrical pulse to the motor, therefore the motor steps one tooth. Through the gearing and here a ball screw the trigger advances say 0.00001". So 10 pulses is 0.0001" and 100 pulses is 0.001". The displacement sensor is just a counting of pulses. Now the problem. The entire chain has a stiffness (its a spring) and compresses with the increasing load applied to the trigger. So at the start (low load) 100 pulses is 0.001", as load increases then next 100 pulses is 0.00095", The next 100 pulses is 0.00090, etc. etc. If the drive system stiffness isn't on the order of more then 100x of the trigger system stiffness then get a low number for actual trigger movement.

The issue isn't with the stepper motor. Its with the lazy way the designer used it to measure displacement.

As you noted, there will be a discrepency between the actuator position and the sear position. This can be compensated for if you can determine the load cell spring rate. It should be linear. Your values of sliding friction would be slightly displaced along the x axis (in a graph) and by an easily calculated amount.
The loadcell stiffness is just one element in the entire drive system. Generally one seeks an estimate of the trigger system stiffness. Then designs a drive system to be >100x of that. But one is never sure about the drive stiffness so one several loads on the drive system to measure its stiffness. However this entire mess can be bypassed by not placing the displacement sensor in the load path. Now the load on the sensor never changes and good results are obtained.

We are going to get a lot of caterwauling about crazy impractical engineers over this post....
 

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I find it so telling that the OP has stated earlier (post #168) that the only thing keeping the hammer cocked without the sear spring is friction, when the geometry shows otherwise ...
Is this with or without hammer hook tip overhang?

If the hook tip lands on a 90 degree sear face, the hammer will drop steadily throughout the trigger pull (negative engagement). The sear is trying to escape.

With hook tip overhang, and either ordnance or 90 degree hook angle, the hook can capture the sear. The engagement is positive for the first tiny bit of sear travel.

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Is this with or without hammer hook tip overhang?

If the hook tip lands on a 90 degree sear face, the hammer will drop steadily throughout the trigger pull (negative engagement). The sear is trying to escape.

With hook tip overhang, and either ordnance or 90 degree hook angle, the hook can capture the sear. The engagement is positive for the first tiny bit of sear travel.

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"Friction plays a very large role in keeping the sear under the hook. Without it the over-center force of the hammer on the sear would instantly cam out the sear." QAZ
???
No, the the force is not over-center, but is drawing it into itself.
The hook tips are pushing down on the sear primary and causing it to rotate into it, not "cam it away". It appears that friction plays a greater roll in rotating the sear out from under the hook tips which are aggressively bearing down on the sear primary, regardless of the fact that the spot they are bearing on is .0000X higher than the escape edge. With the tension on the sear it doesn't slip out, but snaps out when enough force is bearing on the sear legs, this .0000X" is important for a flat plane sear primary, but the same event can/does happen with the TR as well. What's ironic is we painstakingly polish these surfaces to reduce friction as much as possible.

611858
611859


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"Friction plays a very large role in keeping the sear under the hook. Without it the over-center force of the hammer on the sear would instantly cam out the sear." QAZ
???
No, the the force is not over-center, but is drawing it into itself.
The hook tips are pushing down on the sear primary and causing it to rotate into it, not "cam it away". It appears that friction plays a greater roll in rotating the sear out from under the hook tips which are aggressively bearing down on the sear primary, regardless of the fact that the spot they are bearing on is .000X higher than the escape edge. With the tension on the sear it doesn't slip out, but snaps out when enough force is bearing on the sear legs, this .000X" is important for a flat plane sear primary, but the same event can/does happen with the TR as well. What's ironic is we painstakingly polish these surfaces to reduce friction as much as possible.

View attachment 611858 View attachment 611859

LOG
The direction of hook tip force is not inward, it is outward, if that's what you are basing that on.

You can trace the hook force direction right on your photos. It lies outside the sear axis.

The fact that the hook resembles a "hook" is also irrelevant, as only the edge of the tip is in contact with the sear face. I realize you didn't mention that; it's just something that can be misconceived.

If the hook tip were somehow holding the sear captured, how would explain the hook tip falling throughout the sear travel? Wouldn't it necessarily have to rise?

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I've posted this several times and have not gotten any response, not a real concern, as most give it a try and say Oh! But you can watch the sear tip as you pull the trigger as slowly as you can and see if it moves before the hammer blocks your view. In really good light you can look down into the action between and see the polished sear tip through the slot in the hammer for the strut, Hold the hammer back and play the trigger and you will see the sear moving in and out. Now that you have visually identified the sear allow the hammer to rest against it in full cock. Pull the trigger slowly and see if it moves. If you see a slight bit of movement it means the sear primary is slightly off and not releasing in a clean/crisp way.

Please lets not forget inertia and the effects it causes as an affect. When the table cloth is pulled out from under a dish does the dish not stand still due to inertia, and then with the table cloth out from under, it drops to the table?

LOG
The direction of hook tip force is not inward, it is outward, if that's what you are basing that on.

You can trace the hook force direction right on your photos. It lies outside the sear axis.

The fact that the hook resembles a "hook" is also irrelevant, as only the edge of the tip is in contact with the sear face. I realize you didn't mention that; it's just something that can be misconceived.

If the hook tip were somehow holding the sear captured, how would explain the hook tip falling throughout the sear travel? Wouldn't it necessarily have to rise?

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You say outside the sear axis, when the contact is obviously between them with the hook tips (quite aware) pressing the sear down, between the hammer and sear axis, it isn't pushing it away. If what you are saying is true the greater force the hammer hooks bear on the sear will push it away when in fact that would ultimately destroy one element or the other. Trace the hook rotation and it is between them. Put the hammer and sear on the outside of your frame and play a little. And no ,I have not said the hammer captures the sear as long hammer hooks do, causing the hammer to further cock as the trigger is pulled.

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If the sear is pushing the hammer up (counter rotating relative to intended direction) then that is a force overcome which was holding the sear into the hook. The hook tips are advanced relative to location of the inside corner of the hooks. This would tend to push the sear inward.

Even using the rotational tangent line of force at the hammer tips at point of contact on the sear you’ll see the sear is at top dead center of its rotation relative to contact with the hammer. This is as good as a solid rod at this position. Anyone doing cylinder leakage tests on engines know a connecting rod position not “straight up” is going to rotate the engine crankshaft when you put the air to the cylinder. Even a couple degrees on either side of TDC is effective because at that point mechanical advantage is negligible. This is true of any rotaional/linear movement relationship.

If the sear is actually at a disadvantage then any force beyond pull force necessary to move the sear off the hooks applied to the hammer spur would cause the hammer to push the sear out. And the hammer spur point of contact is a greater moment arm than the sear leg radius...i.e. greater leverage.
 

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You say outside the sear axis, when the contact is obviously between them with the hook tips (quite aware) pressing the sear down, between the hammer and sear axis, it isn't pushing it away.
I disagree. Look again. Draw a radius from the hammer axis to the hook tip. Then draw a perpendicular to that radius. That is the hook force direction (the force vector). Its direction is outside the sear axis.

If what you are saying is true the greater force the hammer hooks bear on the sear will push it away when in fact that would ultimately destroy one element or the other.
Not true. Both friction and any cam force present are proportional to the hammer hook force. If you increase the hook force, you also increase the friction holding the sear in place. If the friction is sufficient at any hook force, it will be sufficient at all hook forces.

Trace the hook rotation and it is between them. Put the hammer and sear on the outside of your frame and play a little.
The only hook rotation that matters is that which occurs during the tiny disengagement travel. And that rotation of the force vector still places it outside the sear axis.

But all that aside, a simple mechanical principle remains: if the hammer hook tip falls, the engagement is negative, and the sear would move out from under the hook tip in the total absence of friction. (Ok, maybe it would stop right at the escape edge, to be perfectly correct).

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If the sear is pushing the hammer up (counter rotating relative to intended direction) then that is a force overcome which was holding the sear into the hook. The hook tips are advanced relative to location of the inside corner of the hooks. This would tend to push the sear inward.
The sear is only pushed inward by the sear spring. There is nothing in the geometries causing that.

Think about it. If the geometry was forcing the sear inward, the hooks would be rising in the process. Pretty good trick, raising a weight with no external force.

If the sear is actually at a disadvantage then any force beyond pull force necessary to move the sear off the hooks applied to the hammer spur would cause the hammer to push the sear out. And the hammer spur point of contact is a greater moment arm than the sear leg radius...i.e. greater leverage.
The sear is not at a disadvantage; in fact it has quite a large advantage. Friction is more than sufficient to hold the sear in place.

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You have an x-y coordinate system which is fixed to the sear. This coordinate system rotates relative to fixed space. The center of the coordinate system is the sear's center. The x axis runs along the face of the sear where +x is directed to the left and +y is upward.

x_sear is the current location of the hammer hook tip. You decrement x_sear by a fixed amount equal to dx_sear till x_sear=0 at the escape edge. The escape edge is at the 90 deg vertex of the sear. dy_sear is equal to the difference between y_sear_1 and y_sear_2.
Correct, except I rotate the sear, not the coordinate system. I should have been clearer about that.

The illustration below shows the sear as it would be at three different points of analysis. In each case, the sear axis is directly below the hammer hook tip.

The original purpose of that illustration was to show the relative movement of the sear face, but it also shows the orientation used in the cam force analysis.

You are assuming the hypotenuse of your right triangle is a constant 0.4055 for all calculations of y_sear. This is only true when the hammer tip is at x_sear=0.030. The hypotenuse varies with the value for x_sear getting smaller with each smaller value for x_sear. Separately, for a 90 deg sear, y_sear will be constant and dy_sear=0

You are calculating y_sear=sqrt(x_sear^2+ Hyp^2), where Hyp is a constant. So dy_sear/ dx_sear= 1/2(x_sear^2+Hyp^2)^(-1/2)(2x_sear)= x_sear/(x_sear^2+Hyp^2)(1/2).

For dy_sear/ dx_sear= x_sear/(x_sear^2+Hyp^2)(1/2) we looking at the two terms under the sqrt. As x_sear^2<<<Hyp^2 this causes dy_sear= x_sear/Hyp. As x_sear approaches 0 at the escape edge dy_sear also approaches 0.
No. I am calculating the hypotenuse. For each value of x_sear, the hypotenuse is y_sear, the vertical height above the sear axis. This is then also the vertical height of the hook tip above the sear axis. The hook tip does not move in a vertical direction though, hence the alpha correction (more about that below).

The cam force is F_cam= cos(alpha) F_hook dy_hook/dx_sear where dy_hook=dy_sear/cos(alpha).

Upon substitution, F_cam reduces to F_hook (dy_sear/dx_sear) with no dependence on the angle alpha.
Good observation. Yes, the cos(alpha) and 1/cos(alpha) terms do cancel. I left them in there for a couple reasons.

First, I also use the intermediate calculation results to plot hammer hook displacement. So the 1/cos(alpha) correction is needed for the hook displacement values.

More importantly, if I left those terms out, it would not be obvious that the hook force angle was being taken into account.

As x_sear approaches 0 (the escape edge) the value for dy_sear also approaches zero (dx_sear is the non-zero incrementor), this causes F_cam to also approach zero. Yet we know F_cam doesn't go to zero at the escape edge.
Not sure what you mean by F_cam not going to zero at the escape edge?

611863


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Discussion Starter · #294 · (Edited)
I disagree. Look again. Draw a radius from the hammer axis to the hook tip. Then draw a perpendicular to that radius. That is the hook force direction (the force vector). Its direction is outside the sear axis.



Not true. Both friction and any cam force present are proportional to the hammer hook force. If you increase the hook force, you also increase the friction holding the sear in place. If the friction is sufficient at any hook force, it will be sufficient at all hook forces.



The only hook rotation that matters is that which occurs during the tiny disengagement travel. And that rotation of the force vector still places it outside the sear axis.

But all that aside, a simple mechanical principle remains: if the hammer hook tip falls, the engagement is negative, and the sear would move out from under the hook tip in the total absence of friction. (Ok, maybe it would stop right at the escape edge, to be perfectly correct).

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Yes... This is what is occurring.

I would only add that while forces on the sear would balance right at the escape edge, the momentum the sear developed as it slipped toward the escape edge would carry the sear beyond its edge and the hammer would be released.

I think I may now understand the reasoning by some that the sear "snaps-out" as the trigger is pulled.

I think we (or at least me) owe forum members images of more then just the sear. We need to show a wide-shot picture of the sear and hammer when at full cock, then successive images zooming in to just the sear face and hook notch. I made a poor assumption that when members stated only the tips of the hammer hook contact the sear's engagement face at hammer touchdown (ie negative engagement, ie hammer falls with sear rotation) they fully saw what this geometry establishes in terms of forces and their directions on the sear. This was a mistake on my part.

I'll be out-of-pocket for awhile and won't be able to create these pictures for a couple of days.
 

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Discussion Starter · #295 ·
Correct, except I rotate the sear, not the coordinate system. I should have been clearer about that.

The illustration below shows the sear as it would be at three different points of analysis. In each case, the sear axis is directly below the hammer hook tip.

The original purpose of that illustration was to show the relative movement of the sear face, but it also shows the orientation used in the cam force analysis.



No. I am calculating the hypotenuse. For each value of x_sear, the hypotenuse is y_sear, the vertical height above the sear axis. This is then also the vertical height of the hook tip above the sear axis. The hook tip does not move in a vertical direction though, hence the alpha correction (more about that below).



Good observation. Yes, the cos(alpha) and 1/cos(alpha) terms do cancel. I left them in there for a couple reasons.

First, I also use the intermediate calculation results to plot hammer hook displacement. So the 1/cos(alpha) correction is needed for the hook displacement values.

More importantly, if I left those terms out, it would not be obvious that the hook force angle was being taken into account.



Not sure what you mean by F_cam not going to zero at the escape edge?

View attachment 611863

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Okay. Thanks for the clarification. I need to put in the appropriate amount of time your analysis deserves. I don't have that time now and will respond in a couple of days.

So that I'm thinking correctly, your x-y coordinate system is fixed in space with sear pin center the origin. While it is fixed in space it is rotated relative to the the axis of the barrel. The rotation is such that your y-axis points from sear pin center to hammer hook tip. Your dx_sear, which is the fixed incrementor, may not produce a constant displacement of the of the hook tip along the sear's engagement face due to this rotation. That is, if the engagement is 0.020" the sum of all the dx_sear increments will not equal 0.020" when the sear escape edge reaches the hook tips.

Also I keep forgetting that F_cam is not the hammer force.
 

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I would appreciate an opinion from the masters of engineering. Below are three figures. In each figure there is a block (blue rectangle), and applied force vector (red arrow), and resultant force vector(s) (blue arrow(s)). The blue outlined block is floating motionless (relative to the applied force) in free space.

611869


In figure 1, the top surface block is oriented perpendicular to the applied force vector. It seems to me that the block would be accelerated in the same direction as the applied force vector, moving neither towards the left or towards the right.

611870


In figure two, the top surface of the block is rotated clockwise to the applied force vector. Here, I assume that the block will be accelerated in the direction of the applied force vector as well as to the left as viewed.




611871


In figure three, the top surface of the block is rotated counterclockwise to the applied force vector. Here, I assume that the block will be accelerated in the direction of the applied force vector as well as to the right as viewed.

Please let me know if these assumptions are correct, or not.
 

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We have met two members that have nothing to share in regards to gunsmithing.

Answer not a fool according to his folly, lest you be like him yourself.

I have failed, and apologize to the membership as I did not follow the path,

and failed to answer a fool according to his folly, lest he be wise in his own eyes.

For my penance I promise to never engage with either of them again, lest I become one in the same.

LOG
 
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We have met two members that have nothing to share in regards to gunsmithing.

Answer not a fool according to his folly, lest you be like him yourself.

I have failed, and apologize to the membership as I did not follow the path,

and failed to answer a fool according to his folly, lest he be wise in his own eyes.

For my penance I promise to never engage with either of them again, lest I become one in the same.

LOG
As Don Quixote said, " Lemme take a shot at that windmill!"
 

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Since my conclusion in Post #296 has not been opposed, I will continue:

In the drawings below, the hammer and sear are pictured relative to one another as they would be engaged at full cock in a 1911 pistol having all parts of nominal dimension according to ordnance prints with two exceptions.

  • The hammer hooks have been shortened to about .020”
  • The sear face has been relieved by removing about .010” of its tip at an angle of about -45° to the sear face.
Figure 1 illustrates (red circle) the path of travel of the hammer hooks as the hammer is rotated into and out of engagement with the sear:

611888

The red circle above also represents the direction of force from the mainspring as delivered by the hammer hooks. At the point of full engagement with the sear, this force can be represented by a vector with a magnitude of about 10 lbs and a direction of about 114.5° (these numbers are for illustration only, please choose those that suit you) with that vector being represented by the red arrow in figure 2:

611889


The hammer force vector is acting upon the sear face. The sear face was cut at 90° to a line drawn through the center of the sear pin hole and the sear face rear tip prior to the relief cut being made. The sear has been rotated clockwise about 17° from vertical to fully engage the hammer, so the sear face is at an angle of about -17° to the horizontal as indicated by the blue line in figure 3:

611890


The resulting internal angle between the hammer force vector and the sear face is about 82.5° as shown in figure 4:

611891


Since this angle is less than 90°, the hammer force will be trying to move the sear face to the right. The magnitude of the resultant force to the right will be about COS(82.5°)*10lbs = ~1.3Lbs. This force represents a third force acting on the sear. It will aid the force due to friction and oppose the moment of rotation created by the hammer vector passing to the left of the sear pin center. If the sear face were cut such that the sear face were at an angle of about -24.5° at engagement, the hammer force vector would be perpendicular to the sear face and the sear would be acted on only by the force of friction and moment of rotation created by the hammer vector passing to the left of the sear pin center. If the sear face were cut such that the sear face were more negative than -24.5° at full engagement with the hammer, the hammer force would be trying to move the sear face to the left, out of engagement.

Adding up the forces:

  • Torque due to 10lb vector @ -114.5° with moment arm of .074” off sear pin hole center = 10lb * .074” = .74 in lb. Dividing .74in lb by .4045” sear nose length gives -1.83 lbs of force to rotate the sear nose CCW.
  • Force due to angle of sear face to hammer force of 82.5° = 10lb *COS(82.5°) = 1.31 lbs of force to rotate the sear nose CW.
  • Friction between hammer and sear with coefficient of friction of .12 (pick your own) = 10lb * sin(82.5°) = .66 lb resisting sear CCW rotation.

Total force resisting sear nose rotation out of engagement = -1.83lb + 1.31lb + 1.19lb = .67LB = 11 oz plus any other friction (ie: part pin hole to pin to receiver pin hole).
 

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Since my conclusion in Post #296 has not been opposed, I will continue:

In the drawings below, the hammer and sear are pictured relative to one another as they would be engaged at full cock in a 1911 pistol having all parts of nominal dimension according to ordnance prints with two exceptions.

  • The hammer hooks have been shortened to about .020”
  • The sear face has been relieved by removing about .010” of its tip at an angle of about -45° to the sear face.
Figure 1 illustrates (red circle) the path of travel of the hammer hooks as the hammer is rotated into and out of engagement with the sear:

View attachment 611888
The red circle above also represents the direction of force from the mainspring as delivered by the hammer hooks. At the point of full engagement with the sear, this force can be represented by a vector with a magnitude of about 10 lbs and a direction of about 114.5° (these numbers are for illustration only, please choose those that suit you) with that vector being represented by the red arrow in figure 2:

View attachment 611889

The hammer force vector is acting upon the sear face. The sear face was cut at 90° to a line drawn through the center of the sear pin hole and the sear face rear tip prior to the relief cut being made. The sear has been rotated clockwise about 17° from vertical to fully engage the hammer, so the sear face is at an angle of about -17° to the horizontal as indicated by the blue line in figure 3:

View attachment 611890

The resulting internal angle between the hammer force vector and the sear face is about 82.5° as shown in figure 4:

View attachment 611891

Since this angle is less than 90°, the hammer force will be trying to move the sear face to the right. The magnitude of the resultant force to the right will be about COS(82.5°)*10lbs = ~1.3Lbs. This force represents a third force acting on the sear. It will aid the force due to friction and oppose the moment of rotation created by the hammer vector passing to the left of the sear pin center. If the sear face were cut such that the sear face were at an angle of about -24.5° at engagement, the hammer force vector would be perpendicular to the sear face and the sear would be acted on only by the force of friction and moment of rotation created by the hammer vector passing to the left of the sear pin center. If the sear face were cut such that the sear face were more negative than -24.5° at full engagement with the hammer, the hammer force would be trying to move the sear face to the left, out of engagement.

Adding up the forces:

  • Torque due to 10lb vector @ -114.5° with moment arm of .074” off sear pin hole center = 10lb * .074” = .74 in lb. Dividing .74in lb by .4045” sear nose length gives -1.83 lbs of force to rotate the sear nose CCW.
  • Force due to angle of sear face to hammer force of 82.5° = 10lb *COS(82.5°) = 1.31 lbs of force to rotate the sear nose CW.
  • Friction between hammer and sear with coefficient of friction of .12 (pick your own) = 10lb * sin(82.5°) = .66 lb resisting sear CCW rotation.

Total force resisting sear nose rotation out of engagement = -1.83lb + 1.31lb + 1.19lb = .67LB = 11 oz plus any other friction (ie: part pin hole to pin to receiver pin hole).
You may want to take another shot at that windmill.

Your math is wrong.

You specified the force due to angle of sear face to hammer force of 82.5° as 1.31 lbs to rotate the sear nose CW.

In the total, you have added this force. It should be negative, as it aids in rotating the sear CCW, out of engagement.

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