heavier bullet= less powder?????

[660grizzlyguy]

If I want to change from a 150gr. bullet to a 180gr., keeping the powder type the same, why do I use less powder than my previous load? I would think that since the heavier bullet will take more energy(work) to launch, that more powder would be needed. The load manuals don't see it that way though. Help me see the light.

[azredhawk44]

I would think that since the heavier bullet will take more energy(work) to launch, that more powder would be needed. The load manuals don't see it that way though. Help me see the light.

Energy = Mass * Velocity^2

You just increased mass.

The gun can only contain so much energy before it becomes a hand grenade (ouch!) rather than a bullet launcher.

So, you have to reduce velocity so that the total energy calculation is the same due to the increased mass.

Thus, you reduce powder.

Also... a heavy bullet has more resistance in the barrel. This builds pressure more quickly than a lighter bullet. Having the same amount of pressure trying to push it out will result in increased backpressure, which gives greater chance of the nasty hand grenade situation I mentioned earlier.

I thought the same as you when I started reloading, but once I learned a bit more it made sense to reduce charges when increasing bullet weight for a given cartridge.

[teeroux]

Hevier bullets are harder to get moving down a barrel than lighter ones. Since hevier bullets are more resistant to moving more pressure builds than a lighter bullet with the same charge.

So powder charges reduce as the bullet weight increases but the pressure is kept in the approximate equal range across all max loading. If you were to load a heavy bullet to the charge of a lighter one it would likely build up pressure to a catostrohic level.

[Christchild]

AZRedHawk pretty much summed it up......

My way of thinking, is a bit more "UNeducated", I guess...

Same Cartridge, Same Powder, more bullet weight, more bullet length. Smokeless Powder ignites, producing gas pressure. Heavier bullets have more......weight.

Heavier bullets have greater SD....Sectional Density.....more length....more friction.

More friction and more weight will cause the gases to build more pressure THAN with the same powder behind a lighter/less surface area bullet...So, use less powder. That keeps Your pressures more safe.

ALOT of the aspects of Handloading ammunition are safety, with Pressures in Mind.

This is just my opinion and how I see things happening in my "Mind-Picture". While it makes sense to me, it may not actually be 100% correct.

[Jim243]

It's the pressure not the powder that moves the bullet. That doesn't sound right but it is correct. Heavier bullet less powder.

Jim

[Shoney]

In essence azredhawk44's post is correct, however, we are dealing with gas/pressure laws which should be expressed as

pV=nRT

where

p is the absolute pressure [Pa],
V is the volume [m3] of the vessel containing the moles of gas, here it is the volume you allow in the shell case, determined by how deep the bullet is seated
n is the amount of substance of gas [mol], here it is in solid form as powder
R is the gas constant [8.314 472 m3•Pa•K−1•mol−1],
T is the temperature in kelvin [K].

This is a gross oversimplification, because as the bullet moves down the barrel, the volume and pressure are constantly changing.

In a simple form, think of it as:
Pressure in the case times Volume of case = 2 Irrelevant numbers RT (for our purposes) times the amount of gas n as being in solid form as powder until the "big bang";
1. the gas constant R remains the same;
2. the temperature T remains constant;
For practical purposes the equation should read PV=n

When you substitute a heavier bullet with a charge meant for a lighter bullet, you are reducing the volume in the case. If you leave the exact same amount of powder in the case with this reduced volume, the pressure must go up, most likely to a dangerous level. To exprress this in equation form, doing the math (remember, the two sides of the equation must balance as in the E=MV X V)

double the pressue for a 1/2 reduction in volume, or 2P X V/2 = n or PV=n

There!!! Simple as pie are round.

[javven]

OK if you're changing something in a load, scrap the idea of using anything else in the load - except maybe the same case

If you like a given powder, great - stay with it, but stay within published load data. Especially in 7mm and 308 where there are tremendous differences in bullet weights available, one powder is unlikely to cover all bases. A given powder might be ideal on the light side but dangerous with heavier bullets.

Some calibers like certain powders nearly across the board but a 30 grain bullet change will most likely result in another powder being more accurate.

[David Wile]

Hey Shoney,

I could have answered the Grizzly Guys question, and it would have been pretty much as the others had answered before you. I learned a long time ago about the inverse relationship of bullet weight and powder charge for a given caliber, and it made sense to me as soon as I thought about it for a minute.

Then you come along with your response that reads like something Robert Oppenheimer's worked on with the Manhattan Project. You know, I have not lived a really sheltered life. I went to school, did a little bit of Algebra and Geometry, and I even graduated from a real college with a real Bachelor's degree and all. I know I am not the brightest person around our parts, but I figured I was at least in the middle of the herd.

Now I feel really bad after reading your post. Notice I said reading your post - not understanding it - not one bit of it. For some time now, I thought you were just another old fellow like myself with a little bit of reloading knowledge gained over the past 50 years. Now I can see I was wrong. Either you are one really smart fellow, or else all that stuff you wrote you copied out of some of the Manhattan Project reports. You sure know how to make a guy feel humble.

Best wishes,
Dave Wile

[Shoney]

Dave, Dave, Dave

It ain’t physucks, it’s better living thru kemistree.

Since you have a solid grasp on inverse relationships, it should be as ponderously plain as a pachyderm’s proboscis.

The inverse relationship is between the pressure and the volume in a case just after the Screwey Lewie and the Big Kablooie, you know the “Little Bang Theory” that comes just after the trigger is pulled. Not unlike the big whinger when you pull someones finger. Oops! Sorry, I got sidetracked there.

When you have worked out a max load with a 150 gr bullet, you have a set amount of powder in a case. If you decrease the volume in that case by seating the 150 gr bullet deeper into a case, you increase the pressure when fired.

Likewise, if you use that set amount of powder for the 150 gr bullet and decrease the volume of the case by seating a heavier longer bullet into the case (since the OAL must remain roughly the same, more of the heavier bullet will be inside the case), the pressure will go up at firing. That is why the amount of powder (n, the gas in solid form) must be decrease with the heavier bullet. PV=nRT

Do you suppose that humble pie are square?????? Dave???? You know I took off my alchemists robes and pointed hat to write this, and now my pointed head is cold.

[Nnobby45]

The 180 gr. bullet creates more resistance than the 150 and confines the powder more--- which makes it burn faster and create more energy. An increase in bullet weight requires a decrease in powder charge.

Don't think there's a mathematical formula that's precise. Loads for bullet/powder combinations are worked out in a laboratory and the data is published. As a GENERAL rule, if you increase the bullet weight by a certain %, pressure increases similarly.

Or, if you increase the powder charge by a % with the same bullet, the pressure increase would likely (but I won't say always) be similar.

[SL1]

I hate to throw cold water on Shoney's theory about pressure vs bullet weight, but the "perfect gas law" [PV = nRT] does NOT adequately predict pressure for handloads, SO, PLEASE DON'T TRY TO USE IT TO ADJUST LOAD DATA FOR ONE BULLET TO ANOTHER BULLET OF A DIFFERENT WEIGHT, because it would UNDERESTIMATE the pressure for heavier bullets.

The main reason that the prefect gas law doesn't work here is that smokeless powder burns faster as the pressure increases. So, if you try to compute the pressure for a particular volume, any increase in pressure from reducing the volume ALSO produces MORE GAS by making the powder burn faster and produce gas faster. [In effect, the "n" in the equation is NOT a constant when the gas is being created during the period being covered by the calculation.]

To try to make a short answer to the original question: Typically, a heavier bullet is seated deeper in the case and leaves less room for the powder. But, even if you reduced the powder charge in the same proportion as the space for it is reduced, the pressue would still be higher. The reason is that the peak pressure does not occur until the bullet has moved some down the barrel. So, the amount that the bullet moves before the peak pressure is reached is an important factor, and heavier bullets accelerate slower than light ones WITH THE SAME PRESSURE pushing it. And it is the gas pressure that is limited (by the gun's strength), so the loads are trying to achieve about the SAME PRESSURE. Therefore, the powder charge actually needs to be adjusted such that the pressure will be roughly the same for whatever distance the bullet has moved for each bullet weight. The fact that the powder burns faster at higher pressure makes this adjustment much more "touchy" than the perfect gas law would indicate.

There are computer programs available to handloaders that can calculate loads, based on desired peak pressure, case volume, bullet weight, bore diameter, etc. I think the best one is QuickLOAD, which does a simulation of the power burning process. Others include "Load from a Disk" that uses correlations originally developed by Homer Powley and published in the 1960s by the NRA. A look at those correlations will quickly show you how different they are from the perfect gas law.

SL1

[javven]

Or.... follow load data and work up maximum loads carefully.

[teeroux]

delete

[Shoney]

SL1

Just curious. Did you overlook my statement in my first post on this subject? Where I stated:

pV=nRT

where

p is the absolute pressure [Pa],
V is the volume [m3] of the vessel containing the moles of gas, here it is the volume you allow in the shell case, determined by how deep the bullet is seated
n is the amount of substance of gas [mol], here it is in solid form as powder
R is the gas constant [8.314 472 m3•Pa•K−1•mol−1],
T is the temperature in kelvin [K].

This is a gross oversimplification, because as the bullet moves down the barrel, the volume and pressure are constantly changing.

There was no attempt to predict pressures, only show in very simple mathmatical terms, why the powder charge is lowered for heavier weight bullets.

If as you say:

The main reason that the prefect gas law doesn't work here is - - -

Then pray tell, what laws are used to compute this?????

If it is not a complex form of the pV=nRT; taking into accout the changing pressures and changing volumes and changing amounts of gas at changing temperatures???? What is the basic formula that is used??????

[DaveBeal]

Shoney, I don't think SL1 was arguing or criticizing. He was just adding detail.

PV = nRT is the fundamental truth here, but SL1 is correct in explicitly noting that "n" is not constant, regardless of whether a light or heavy bullet is used; it increases as the powder burns. And as he also mentioned, there's a double whammy in using a heavier bullet: (1) Being heavier, it's slower to accelerate and so confines the gas for a longer time (smaller "V"), which increases pressure, and (2) the increased pressure causes the powder to burn more quickly, which causes "n" to increase more rapidly than with the light bullet, increasing the pressure even more.

[David Wile]

Hey folks,

Like I said to Shoney, I have no chance of figuring the whole pV=nRT thing. Kind of like the E = MC2 thing, I've heard of it, kind of know what it means, but I could never do anything with the formula.

However, I do understand a concept that seems to make sense to me as to why a heavier bullet will call for a lighter charge of powder. Grizzly started the thread by comparing the use of a 150 grain bullet and a 180 grain bullet and using the same type of powder for each. He wonders why you would not use more of the same powder when using the heavier bullet. I think I can set up an example that will make sense and be easy to understand, and none of us dumb folks will have to use any formulas. We just have to think about it a bit in our minds, and it should make sense on our lower levels.

Lets assume a few things: we are using the same bolt action rifle for our examples, we are using the same Type XYZ powder for our examples, and the only thing we are changing in our examples is the weight of the two bullets which are 150 grains and 180 grains and both the same type and design of jacketed bullets.

Now, first we start with the 150 grain bullet. It has a maximum charge of 50 grains of Type XYZ powder behind it, and it produces a maximum pressure which is just a hair bit under whatever pressure the bolt action rifle is designed to take without coming apart. In other words, if we shoot our 150 grain bullet with 50 grains of powder, the action will contain the pressure. However, if we put 51 grains of powder behind that 150 grain bullet, the pressure will be increased and rupture the action of the rifle. The extra 1 grain of powder produced more gas and pushed the pressure past the capability of the action.

Now, let's back things up one shot. We fired our first shot of 50 grains of powder behind the 150 grain bullet, and the rifle action held together. We now decide we want to put a 180 grain bullet over that 50 grains of powder instead of the 150 grain bullet. We are not increasing the powder, so we cannot be increasing the energy supplied by the burning powder. So what is the problem?

The problem is this. While the energy released by the burning powder may be the same, the heavier bullet causes a significant increase in chamber pressure because of inertia. The 180 grain bullet is more difficult to start moving and keep moving than the 150 grain bullet. This causes the significant increase in chamber pressure, and it causes the action to rupture. Remember, the 150 grain bullet with 50 grains of powder created a chamber pressure just a hair bit under the rupture level, and any further increase in chamber pressure would result in a rupture.

Does that scenario mean a 180 grain bullet cannot be used in that rifle? No. A 180 grain bullet can be used if you do something to reduce the chamber pressure below the rupture point of the action. In our imaginary case here, we can reduce the peak chamber pressure for the 180 grain bullet by reducing the powder charge by some given amount.

Companies like Lyman and others test fire different bullet weights with different powder charges in special firearms that are designed to measure pressures produced, and they come up with safe loading data that they publish in the manuals they sell. I hope my imaginary scenario helps some other less scientific folks like myself to understand why we use less powder for heavier bullets as Grizzly was asking, and I think you will find all published data in any given manual will show the same thing.

How did I do, Shoney?

Best wishes,
Dave Wile

[Shoney]

Dave:

Yup! By jove, you've got it. When I was reading that, the room got blinding bright, harps were playing, Angels were beautifully singing "Brevity is the sole of wit."

OK! I'll shut up and drimk my prune juice and Burgandy.

[HiBC]

The fly in the ointment is using the same powder(maybe)

If you are using an ideal powder for a 150 gr bullet,it will be efficient for accellerating your 150 gr bullet with its burn rate.If you put a heavier bullet,the accelerates slower,in front of the powder,the pressure will build quicker.
We are balancing pressure,time,and the strength of the gun,the barrel length,etc,

Think dragster.Compression ratios,tires,gearing

Generally,what I do,instead of trying to back off a powder that is too quick,I go to a slower powder in a heavier bullet.Its like changing gearing for a heavier load.I try(within the limits of recomended safe loads) to use a powder that lets me fill the case up most of the way.
The barrel is the length of the racetrack.I like to be still accelerating at the end of the trtack.

[SCOTTYDOG]

GrizzlyGuy; Are you completely confused yet?

[SL1]

You asked

Just curious. Did you overlook my statement in my first post on this subject?

No, I saw that, and I wasn't trying to be critical of you. But, I was concerned that somebody who read it would try to USE it to do some "modifications" to data that they had in order to load a bullet for which they did not have data. That is someting that probably every one of us has done at some time, using various techniques to "guestimate" a load for which we have no data.

You also asked

Then pray tell, what laws are used to compute this?????

The accurate answer is that you need to use equations that cover
1. the burning rates for powder at various pressures,
2. the DYNAMICS of gases as they flow, including changes of pressure and temperature with velocity, and
3. the dynamics of solid objects (how they are accelerated by forces).

Obviously, I can't write all that down here, and frankly, NOBODY does these calculations from the first-principal equations, anyway. Even QuickLOAD uses some correlations for some of its dymamincs calculations.

But, for those who are interested, the correlations originally developed by Homer Powley were published in 1981 by the NRA in a book written by William C. Davis, Jr., titled "Handloading". It is in a 7-page chapter that includes tables, so it is not feasible for me to type out enough information here to make it useful to somebody. But, it doesn't require any more math than knowing how to raise numbers to powers (e.g., the square, cube and 4th power of a number) and take roots of numbers to powers (e.g., the square root, cube root and 4th root). Today, most calculators can do that for you. For folks who decide to get a copy, be aware that there are 2 typos in the article that do become apparent if you study it and follow-along with the examples.

For those of you who like to think about these things, I'll give you something to think about with respect to trying to use the perfect gas law. We all know that velocity increases approximately in proportion to the charge increase as we add powder to a given bullet/case combination. (That is, a 1% increase in powder gives about a 1% increase in velocity.) And, we are told, that pressure increases about twice as fast as velocity, so that means that the pressure is increasing approximately as the square of the powder addition.

But, adding powder adds energy in direct proportion to the amount of powder added. So, you might think that adding powder would increase the bullets muzzle energy in proportion to the charge increase (by assuming that the gun's efficiency for converting powder energy to muzzle energy doesn't change much). But, that would increase velocity by only one-half the fraction of the charge weight increase, because velocity is proportional to the SQUARE ROOT of the muzzle energy. Thus, this way of thinking aobut it would predict only half as much increase in velocity as we actually see. (For some published numbers to verify the relationship between charge weight, muzzle velocity and peak pressure, see Richard Lee's book "Modern Reloading" and look at the "one grain factors" in his data and the text on pages 124-128.)

So, how does the velocity increase TWICE AS MUCH as the increase in propellant energy would seem to predict? If you use the perfect gas law with an increase in "n" to account for the increase in molecules in the gas, then what do you do about the temperature, "T"? (It is NOT constant, like Shoney assumed in his first post.) You might try to think about that by thinking about an example where the case volume increases by the same fraction as the charge weight, then think about compressing the gas back to the original case volume. That would be PV=nRT and P(V x 1.01) = (n x 1.01)RT for the original and the proportionally increased case volume and charge weight, because the powder should have the same conditions in both situations, there would just be more volume with the same conditions, including the same "T" in the two situations. So, to decrease the volume on the left side of the second equation back to V, you could increase the pressure by 1.01 and get P(1.01)V = n(1.01)RT. But, what you would be missing is that it would TAKE WORK (i.e., ENERGY) TO COMPRESS THE GAS BACK TO THE ORIGINAL "V", and that work energy increases the temperature, too. And, the increase in temperature increases the pressure along with it. So, if we think about the relationship between P and T as follows: P(V x 1.01) = nR(T x 1.01) is equivalent to (P x 1.01)V = nR(T x 1.01), then we can see that the energy necessary to increase P and decrease V gives a proportional increase in T. SO, that means for our example that compressing the extra gases into the original volume raises both "n" and "T" by the ratio of the charge increase, so we would get (P x 1.10 x 1.01)V = (n x 1.01)R (T x 1.01). This is telling us that pressure needs to go up by 1.01 x 1.01 = 1.0201, or about twice the fractional charge increase (i.e., 0.02 compared to 0.01) when we increase the charge by 1.01, or 1%. That is ROUGHLY like the data indicates, and it answers the question about how adding 1% more energy in the powder charge produces more than 1% increase in the bullet's muzzle energy - - the pressure increases by a larger factor than the charge weight, and that increases the ballistic efficiency of the load.

But, all of that assumes that the same fraction of the power burns in the same amount of time no matter how much powder is in the same case volume. That is NOT true. As pressure goes up, so does the buring rate and thus the "n" in the equation. But, higher pressure pushes the bullet harder, makes it accelerate faster, and thus makes the volume from the base of the bullet to the back of the case increase faster. That tends to keep the pressure from increasing as fast as it would have if the bullet always accelerated at the same rate, which tends to keep the powder from burning so much faster. Thus, there are counteracting effects that do not balance each other perfectly, and need to be accounted for with detailed calculations. Having done some of those, I can tell you that the net effect is that pressure often increases something like the 2.7 power of the charge fractional increase instead of the 2.0 power that we would get from manipulating the perfect gas law like we just did. But, don't take that 2.7 factor as a useful value for making guestimates, because it varies a lot from that example, depending on the particulars of the load.

That is enough, because both my fingers and brain are tired, now.

SL1

[Mello2u]

660grizzlyguy

heavier bullet= less powder?????
If I want to change from a 150gr. bullet to a 180gr., keeping the powder type the same, why do I use less powder than my previous load? I would think that since the heavier bullet will take more energy(work) to launch, that more powder would be needed. The load manuals don't see it that way though. Help me see the light.

To summarize, there are many factors which effect pressure in reloading. Friction, mass (inertia), and volume are three major factors.

Perhaps the most difficult to understand is the effect of reducing volume by seating a heavier and longer bullet to achieve the correct maximum overall cartridge length. As posted above, reducing volume effects the burning rate of the the smokeless powder by increasing the rate and therefore increasing pressure faster. To avoid dangerous pressures one must reduce the powder charge.

[brickeyee]

If it is not a complex form of the pV=nRT; taking into accout the changing pressures and changing volumes and changing amounts of gas at changing temperatures???? What is the basic formula that is used??????

It is a dynamical system with no closed form expression that models it.

It is done by evaluating the systems conditions at small time increments based on the previous time increments.
Make the ties step small enough and you can get a smooth simulation.

Just about every real gas departs from the ideal gas law to some degree.

Separate gas constants are available for each gas to try and extend the conditions under which the ideal gas law applies.

It still breaks down if any chemical reactions occur in the gas, or if the pressure gets very high.

The ideal gas law also fails of any liquid gas is present.

Go to extremes of temperature and pressure and it has increasing error.\

The gas is no longer 'ideal.'