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Paul Lakowski

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Posts posted by Paul Lakowski

  1. I read that in testing modern APFSDS, the MV can change as much as 40m/s if the temperature is changed by as little as 5°C. THis is the reason that modern tanks have heaters in the tanks, to keep the ammo warm.

    But what about WW-II ammo or ATGunners...how much difference did ammo temperature make on Muzzle Velocity?

    Was there any thing gunners did to keep the ammo warm?

  2. Originally posted by Sergei:

    </font><blockquote>quote:</font><hr />Originally posted by Paul Lakowski:

    This tale is typical of the way some war correspondance is reported. Early in Barbarossa the soviets had little chance of victory so every desprite tale became a focal point to make up for the lack of any good news.

    Even if Raus is lying, it would seem odd for him to resort to Soviet propaganda. So, what kind of person is he anyway? Maybe some analysis of his person and his history would help to understand why he wrote such things after the war. </font>
  3. This tale is typical of the way some war correspondance is reported. Early in Barbarossa the soviets had little chance of victory so every desprite tale became a focal point to make up for the lack of any good news. FOr example the divisions that were allegedly held up actually were to the east of the river encircling soviet mechanized korps destroying 200 tanks for the loss of 40 of there own tanks.

    You see the samething happen to the germans at the end of the war. Great attention is paid to detailed accounts of individual acts of heroism and success while their own army crumbles around them.

  4. Originally posted by Andreas:

    Originally posted by Andreas:

    3. Degradation of the Russian armour steel was not noticeable. However, the armor steel is darker and finished rougher. The tanks reveal they were produced in a short time, because there is no evidence of any close tolerance work. The turret of the T34 is not made form a single piece,; instead it is assembled from numerous pieces. In many T34 tanks the armor walls wre created from pieces of 1cm thick steel with 6cm filling of cast iron or other material and then a second piece of 1cm thick steel.

    What part of the T-34s armor is 8cm thick and what other materials do they use other than steel and cast?
  5. Originally posted by Jeff Duquette:

    </font><blockquote>quote:</font><hr />Originally posted by Paul Lakowski:

    </font><blockquote>quote:</font><hr />Originally posted by Jeff Duquette:

    I’m not sure I get it. You are now saying the velocity component is not included in the spaced effect modifier?

    So the spaced effect, or spaced modifier is not:

    (t/d)^0.5 x 0.1d x V{in Km/s} x gap modifier

    ???

    It should just be Spaced plate erosion by determine the t/d ; angle & effectiveness of the plate material .

    Then (t/d)^0.5 to determine the 'spaced plate effect'

    So 2cm Aluminum [0.5d] vs 3cm API becomes

    2/3÷2^2=0.8 x 2= 1.6

    1.6 x 0.5[te]= 0.8cm

    The spaced plate effect becomes

    0.8/3^0.5 = 0.5cm

    If there is insufficent gap, multiply 'spaced plate effect'by x 0.7.

    If the penetrator is high strength reduce the 'spaced plate effect' by x 0.7

    So if the penetrator was high strength and the gap insufficent it should be x 0.5. </font>

  6. Originally posted by Jeff Duquette:

    I’m not sure I get it. You are now saying the velocity component is not included in the spaced effect modifier?

    So the spaced effect, or spaced modifier is not:

    (t/d)^0.5 x 0.1d x V{in Km/s} x gap modifier

    ???

    It should just be Spaced plate erosion by determine the t/d ; angle & effectiveness of the plate material .

    Then (t/d)^0.5 to determine the 'spaced plate effect'

    So 2cm Aluminum [0.5d] vs 3cm API becomes

    2/3÷2^2=0.8 x 2= 1.6

    1.6 x 0.5[te]= 0.8cm

    The spaced plate effect becomes

    0.8/3^0.5 = 0.5cm

    If there is insufficent gap, multiply 'spaced plate effect'by x 0.7.

    If the penetrator is high strength reduce the 'spaced plate effect' by x 0.7

    So if the penetrator was high strength and the gap insufficent it should be x 0.5.

  7. Hi Jeff I have this paper....

    Originally posted by Jeff Duquette:

    Paul:

    ===============================================

    57mm APC M86

    Comparing your equations with the firing trial results for Armor Arrangement 1:

    57mm APC M86 vs. 2” plate at 40-degrees ---->/

    Firing Trials Ballistic Limit (BL) = 2149-fps => 2149-fps for t/d of 0.89 for APC vs RHA at 40-degrees actually does equate to 2” worth of protection. Checks out.

    =================================================

    Armor Arrangement 2: 57mm APC M86 vs. ½” Skirt Armor 16-inches away from a 2” main plate. Both plates are sloped at 40-degrees and are parallel to each other.

    ---->/ /

    Firing Trials Ballistic Limit (BL) = 1943-fps => 1943-fps at 40-degrees equates to an equivalent single plate of about 1.91”. So the firing trials indicate that this particular spaced armor array is actually less efficient than a single 2” plate inclined at 40-deg.

    Your equation predicts an equivalent armor thickness of about ~3.39”

    Skirt Armor Plate Equivalent = 0.4204”

    Main Armor Plate Equivalent = 2.219”

    Spaced Armor Modifier = +0.747” (at a striking Velocity of 1943-fps or ~0.592Km/s)

    0.4204” + 2.219” + 0.747” = 3.39”

    OK the skirting plate was untreated 1/4 inch plates sandwiched together [doesn't state how].

    This could mean mild steel which can be 180BHN [0.8 Te] or down to ~ 100 BHN [~ 0.66]...anything lower would be soft steel.

    Since its not stated which steel is used this could be either or and will modify the results.

    More importantly is the fact that the researchers identified the 57mm ammo to do better than they thought since it had higher longditudeal properties. Now I'm not sure what they mean't by this but it sounds like high strength [as opposed to high hardness]. Its long been known that high strength penetrators do better than regular strength penetrators in both penetration and multi layered armor penetrations....and spaced armor is no different. THis is why modern APFSDS are moving back to these.

    In modern ballistics tests with APFSDS such a high strength penetrators demonstrate a 70% lower spaced plate effect [goes from 1.3d to 0.9d] . If we apply this to the 57mm case the situation improves dramatically.

    BTW the body diameters of these rounds look to be 5.75cm M70 and 5.6cm M-86.

    Two 1/4 inch mild steel plates and one 2 inch RHA plate workout to [cm sorry I'm sooo European smile.gif ] 0.635cm x 2 plus 5.08cm. The t/d Vs M-86 works out to 0.56 & 0.91 respectively ..adding Te = 0.37 x 1.27= 0.47cm and 4.61cm.

    The spaced plate effect should be SQRT [0.61/5.6] = 0.33d x 0.7 = + 1.3

    The angled plate looks to be 7.9cm or about 3.1 inches. This doesn't explain the parrallel plates but looks close to the opposing plates....

    Looking over the parrallel plates I recall the yaw and pitch factor that altered the impact on the second plate reducing the angle by up to 7°.That never happens with APFSDS so they never added that factor....guess thats a modification that needs to be added for AP level penetrations.

    That should reduce the back plate contribution to 33° or 4.61cm ÷ 0.84= 5.5cm + 1.3+ 0.47cm = 7.3cm or 2.86 inches.

    While this may look like a problem it may not be. The M-86 is an APC. Modern research suggest that the cap will truncate the spaced plate effect....and this may be an indication of just how much.If we eliminate the spaced plate effect due to the cap, then the values become 6.6cm [2.6"] and 6cm[2.36"] respectively .

    It also should be noted that with spaced plate their seems to be a higher scatter on results than normal AP shots with SD being as high as 10%.

    =================================================

    Armor Arrangement 3: 57mm APC M86 vs. ½” Skirt Armor 16-inches away from a 2” main plate. Both plates are sloped at 40-degrees and inclined in opposite directions.

    ---->/ \

    Firing Trials Ballistic Limit (BL) = 2755-fps => 2755-fps 40-degrees equates to an equivalent single plate of about 2.86”. So the firing trials indicate that this particular spaced armor array is actually somewhat more efficient than a single 2” + ½” plate inclined at 40-deg.

    Your equation predicts an equivalent armor thickness of about ~3.71”

    Skirt Armor Plate Equivalent = 0.4204”

    Main Armor Plate Equivalent = 2.219”

    Spaced Armor Modifier = +1.068” (at a striking Velocity of 2775-fps or ~0.845Km/s)

    0.4204” + 2.219” + 1.068” = 3.71”

    I must be gettin old cause I Fu#$ something else!!!

    The spaced plate effect is not modified by V, just t/d...its the back plate effect to determine penetration in the first place , thats modified by V..... big sorry for that :(

  8. Haven't got a clue what Jasons on about so I'll just let it alone for now.....

    Originally posted by Jeff Duquette:

    Thanks Paul. Sorry to be a chuckle-head but I have several more areas of clarification...if you would be kind enough to continue to indulge me smile.gif

    Paul Said: For example 37mm Vs 45mm SHS of T-34 glacis is [45mm/37mm/2 ^0.2]*1.2 or 49mm @ 60°...while if 75mm hits thats

    [45mm/75mm/2 ^0.2]*1.2 or 42mm @ 60°.

    How are you accounting for the angle of attack? Should “Te” be 49mm/COS(60)

    And

    42mm/COS(60) for the 75mm dia projectile attack example?

    Yes usually I don't include the angle since its a property of the projectile penetration...IE if the effective resistances is 49mm @ 60°, then you have to find out at what range 37mm can penetrate 47mm @ 60°...which is none existance.

    So for the 75mm the range at which a 42mm can be penetrated @ 60° should be about 1/2 hits @ 1400m

    =================================

    I am assuming the 1.2 multiplier is a function of the homogeneous hard armor on the T34?

    The RHA multiplier being 1.0;

    FHA multiplier is 1.3; and

    SHS multiplier is 1.2.

    Is this correct?

    Yep with RHA being referenced to BHN 275 rolled plate.

    SHS depends on the amount,450BHn is about 1.25 while 380BHN RHA is going to be around 1.1

    ==================================

    In one of your earlier posts you indicated:

    So 3cm FH @ 10° t/d should become

    (3cm/7.6cm/2)^0.2 x 1.3[Te FH] => 0.723 x 3cm x 1.3 = 2.82cm @ 10°= 2.9cm

    This looks to me like you divided the resultant 2.82cm by COS(10-deg) to get the final thickness of 2.9cm…although I am getting 2.86cm after dividing by COS(10). You rounded up? Was this in fact what you were doing?

    In addition the 50cm Stug-111 plate…shouldn’t the final thickness or “Te” be modified by its angle of inclination as well? 52cm/COS(10)???

    Yes but strickly speaking the angle should be an expression of the projectile.

    And yes to the 52mm FH plate....but usually at shallow angles the COS of the angle is sufficent especially when the projectile is blunt like many soviet WW-II and modern penetrators. When you get into ogive penetrators the projectile will slide and exagerate the LOS penetration through the angled armor.

    [ November 17, 2003, 11:30 PM: Message edited by: Paul Lakowski ]

  9. Originally posted by Jeff Duquette:

    Paul:

    I have a couple of additional questions I am hoping you can wade through. After looking at your write-up a bit closer I realized you are not saying the minimum air-gap would have to be 1.77cm.

    1) What -- if any -- modifications would occur to the spaced armor modification if the plates were some other steel combination…say RHA skirt over RHA main armor instead of the FH over FH which is being considered in this example?

    2) Why is the air-gap advantage applied to the thickness of the skirting plate? Why isn’t the thickness of the main plate employed in determining the spaced armor added advantage?

    3) You have specified an optimal air-gap space of >1d. Is there any added advantage if the air-gap is say 1.5d or 2d or 3d. The way in which you have presented your equation, the spaced armor advantage is purely a function of:

    The skirting plate thickness;

    Projectile diameter; and

    Projectile striking velocity.

    Is there no added advanatge for an air gap wider than 1d?

    4) Along the same lines as Question 3 except in the opposite direction. You have indicated that anything less than about 1d would fall into the range of less than optimum spacing. However you have really not established a lower bound. Is there a point at which no real spaced armor effect would occur. Are we talking 0.1d or 0.5d; or will an air gap equal to the width of an oxygen molecule be sufficient to rate a spaced armor advantage? ;o) Regarding the 90mm HVAP example cited, the warping resulted in a 2-inch airgap. This is about equal to the diameter of the M304's penetrator.

    5) Just to clarify in my own mind…your final assessment of the Stug-III armor you indicated: 2.9cm + 2cm (air gap) + 5.2cm = 8.1cm => 10cm

    The 5.2cm is derived from: ((5cm/7.62cm/2)^0.2)*5cm*1.3….is this correct.

    Thnx

    That is correct.

    For example 37mm Vs 45mm SHS of T-34 glacis is [45mm/37mm/2 ^0.2]*1.2 or 49mm @ 60°...while if 75mm hits thats

    [45mm/75mm/2 ^0.2]*1.2 or 42mm @ 60°.

    # 4) I did wonder about that but the shattering was severe as I recall not minor.In anticipation of that very question I was reviewing some of the ballistics experiments. In one very involved study hard Steel short rods with sharp ogives were test fired into hugh blocks of aluminum. The actual aluminum penetration was easy until it hit the next block placed adjacent to the first. Either the penetrator veered off at 45° after crossing the boundry or it shattered! In similar modern long rod penetration studies against steel the blocks surface of both exited block and reentry blocks were severly distorted showing massive expenditure of energy. Layering in modern ballistic armor is expected to yeild anything from 10-40% improvement in the overall armor resistance. There is a direct connection some where here but I have still not been able to quatify that.

    #3) this is not easy to answer since the damage effect on penetrator is determined by the spaced plate ...and when breaking it down it works best if you make that simple distinction.VEry large gaps tend to increase the overall resistance of long rod penetrators or even short rods @ > 1km/s impact velocity to the tune of 10-20%...but these are over vcery large gaps of 20 diameters. I recall Zukas did a paper that for 10:1 L/d rods @ 1.5km/s the gap increased dramatically when the gap got to 3-4 diameters. On the other hand Holher and Stilp show that below 1km/s impact V, short rods penetrated more than two thinner plates easier than one thicker plate when the air gap was 20 diameters. I tend not to factor large airgaps in.

    #3) where large airgap becomes significant is in the area of shaped charge resistance. The formula is

    The erosion of the skirting plate plus the base armor [minus t/d] and the standoff effect plus 0.7d...so if the skirting plate is 1cm mild steel and the warhead is 10.5cm that adds upto 8.1cm plus the standoff effect[depends on liner material etc].

    2) because its all got to do with how much damage the skirting plate does to the impacting penetrator...remember Dr Elders paper on "spaced armor at sea", if the damage was insufficent to decap the APCBC the projectile could contiue to travel 24 diameters airgap and still penetrate with almost full effect.

    For example the steel mesh the germans employed is perforated plate armor and should increase the damage to the impacting projectile by 1.6 times [and double that if its a HVAP shot]...that explains why those 14.5mm HVAP shattered and did almost no damage to the Pz-III hulls....it literally shatters the penetrator!

    1)If it was straight steel there would be no 1.3 modifier at the end and it would resist less. Additionally if the skirting plate was aluminum its Te is ~ 0.3 to 0.4, so you'd need three times as much thickness skirting plate.

  10. Originally posted by rexford:

    What is the basis for the calculations of face-hardened armor penetration as a function of projectile diameter and plate thickness?

    So what kind of airspace is needed?

    THe basis of all modern ballistics test are semi infinite reference point...this is the only way to compare differing levels of resistance etc. The task is then to back calculate the t/d effect etc.

    Angled impact is controled by projectile nose design, since this controls the back plate effect.

    As to the airgap it doesn't take much to allow shatter to occur, remember the case of the 90mm HVAP shot hitting two plates that were treated as one in the test. Should have penetrated cleanly but ended up shattering inbetween the plates.

    Now that I've had a chance to review this I need to make a correction.The formula was developed from APFSDS back through APDS to AP penetration and in all my modern cals the 'te' is factored independent of t/d so thats in the wrong place in the forumla I put down. It should read...

    [Thickness) ÷ projectile thickness]÷ 2[semi infinite] ^0.2 * Te.

    So 3cm FH @ 10° t/d should become

    3cm÷ 7.6cm ÷ 2^2 x 1.3[Te FH]or 0.723 x 3cm x 1.3 = 2.82cm @ 10°= 2.9cm

    Then 'spaced plate effect' becomes; SQRT [2.9÷7.62] = 0.61* 0.1d*V[km/s]* 0.7[poor gap]= 0.26d or 2cm additional protection.

    That should mean the driver plate offers

    2.9[3cm FH ]+ 2cm[space] + 5.2cm =8.1=>10cm, depending on if the plate bulges or not during impact.

    Jeff 'Te' means 'thickness effectiveness' and the + 2cm is the reduction in penetration due to the airgap. If this air gap was 1d or more [iE 7.6cm] it should result in + 2.8 cm loss in penetration.

    Against higher velocity GErman 75mm guns this type of applique mounted on a KV-1 should mean higher loss due to higher velocity.However with APCBC ammo you have the possiblity that the sectioned nose will truncate the damage done to the penetrator by such spaced plate limiting the loss of penetration...that and the fact that these are longer penetrators which lead to more penetration anyways.

    [ November 17, 2003, 12:15 AM: Message edited by: Paul Lakowski ]

  11. Originally posted by Jeff Duquette:

    Paul said:

    The Stug-IIIF are produced with 30mmFH + 50mm FH armor which resists like 2.6+ 4.8cm or 7.4. However these were bolted and if there is any airgap developed between [warping] the plates that should shatter the projectile. The result is + 0.24d or 1.9cm additional resistance to a total of 9.3cm RHAe, which limits T-34 76mm AP shell to ~0-900m penetration [½ hits], depending on if theres an airgap or not. The Stug-IIIf own 75L46 gun penetrates T-34 at…..

    This is pretty divergent from your post about a year back on the Yahoo Tankers Forum. I had brought up the air-gap being of some significance on the KV-1E’s bolted on armor. You indicated that this small air gap would have little or no effect on normal AP if the dimension of the air gap was not at least as large as the diameter of the projectile. You than went on to say that these smaller air-gaps might affect the performance of HVAP\APCR. Is there some reason why you are now changing your opinion on this particular aspect of terminal ballistics? Bear in mind we are not talking about analogies to long slender APFSDS. These are big fat, blunt projectiles that would be striking a STUG-III’s armor.

    The loss of projectile based on properly spaced plates [>1d] should be 1.3d for monoblock and 2.6d for sheathed . With limited space between plates its minus 0.7d so that becomes 0.8d for monoblock and 1.6d for sheathed penetrators.

    The loss against AP shot should be 0.1d*V[km/s]...if there is sufficent gap [twice for sheathed] and about 70% of this is there is insufficent gap.

    So 76mm @ 600m/s is 0.6d if sufficent gap and 0.42d if not.

    Whats new is the system of calculating t/d. I made a mistake assuming AP lost t/d like APFSDS but its less impact.

    I've figure that for AP shots its more like

    [Thickness* Te) ÷ projectile thickness]÷ 2[semi infinite] ^0.2 .

    So 3cm FH @ 10° t/d becomes

    (3cm x 1.3[Te FH])÷ 7.6cm ÷ 2^2 or 0.76 x 3cm= 2.28cm @ 10°= 2.33cm

    Then 'spaced plate effect' becomes; SQRT [2.33÷7.62] = 0.55* 0.1d*V[km/s]* 0.7[poor gap]= 0.23d or 1.8cm additional protection.

    That should mean the driver plate offers

    2.33[3cm FH ]+ 1.77cm[space] + 4.3cm =8.4cm...which should limit 76mm penetration to ~ muzzle with gap and ~ 1km without.

    I guess I assumed these plates are at 20° angle.Thats why I try to avoid posting such cals...all it takes is one small error ;)

  12. The Stug-IIIF are produced with 30mmFH + 50mm FH armor which resists like 2.6+ 4.8cm or 7.4. However these were bolted and if there is any airgap developed between [warping] the plates that should shatter the projectile. The result is + 0.24d or 1.9cm additional resistance to a total of 9.3cm RHAe, which limits T-34 76mm AP shell to ~0-900m penetration [½ hits], depending on if theres an airgap or not. The Stug-IIIf own 75L46 gun penetrates T-34 at…..

    All hits @ 900m range

    ½ hits @ 1500m range

    ¼ hits @ ~ 2km range

    However later model T-34s the sloping front side turrets employ 52mm and later 65mm cast armor and should reduce penetration range to ~ ½ hits @ muzzle and ¼ hits @ ~ 700-800m range.

    So both statements are true.

    [ November 16, 2003, 02:05 AM: Message edited by: Paul Lakowski ]

  13. Don't mean to rain on the parade here ...especially when most all the points are valid.

    But when the Germans went into Barbarossa about a 1/3 of their tanks were inferior PZ-1 & II tankettes and another 1/4 were either early Pz-III or Czech tanks...all these models had thin armor and 37mm gun, making them at best equal to the bulk of the SOviet tanks [bT & T-26 tanks].

    Thus only about 40% of the German tanks were superior to the main SOviet tanks but the Soviets also had a similar number of T-34 & KV-1s, that were on paper vastly superior to the best german tanks.

    820 pz35t & Pz 38t

    1230 PzI&II and Bef

    1404 PzIII (50) & Pz-IV.

    So looking at the tanks stats is only part of the answer. The fact that the Soviet tanks had only one man turrets and no radio, while the Germans had 3 man turrets and radio...this is important. The germans found they could out shoot the Soviets 3-4:1 and there commanders could better harness this firepower through radio communications.

    By wars end the Soviet tanks all had two to three man turrets and many radios [at least one per platoon].While many German tanks were fielding heavier guns so the shot advantage was down to ~ 2:1 at best...while SU vehicles could often match the German in ROF.

    But the main reason I think is command leadership...the purge removed many soviet commanders that were in the same league as the German generals...a situation that was not really overcome until after the war. Thus they were beat before they even started, the fact that they survived and went on to win the war is by far the most interesting question of the eastern front! Howday do dat?

  14. Any money /production lines spent on enhancing HT is money taken away from tank production...so the key word becomes whats 'adquate' for the task.

    As long as the HT can follow 1km behind the armor and go almost any where they go ...then the mission is accomplished.

    Mind you German SPW in recon battalions was more of a cavalry role and needed more organic mobile fire power...this too was logical because of the way they were employed. One could argue that M-3/5 companies should have taken there 37mm ATG and mounted them on the HT in a fire support role.

    But they wouldn't have been used in the same way the germans did with ATG on there SPWs.

  15. Front turret of the 1940 model of the T-34 was 45mm cast armor . This armor was known to be brittle and suffered when hit by overmatching shells [ wider than the plate thickness]. SO as a rule the 45mm rolled plate glacis @ 60° was in vulnerable to even 50mm rounds but 75mm cut through out to 1500m or more.

    The front turret and glacis did have weak spots that could reduce the armor 10-20% in resistance to a 37mm @ short range through this area [ 2-3 projectile radius = < 11cm or 4 inches for you yankees]. Thats a pretty small area to hit even at 200-300m range. All the battle accounts I've seen of 37mm speak of them repeatedly ricocheting off the T-34 armor.

    I can believe damage could be done but not so many kills?

  16. Its been shown experimentally that when AP type shells penetrate a target the material moves out of the way slower than the striking velocity, thus the tip of the penetrator is always slower than the tail...with modern APFSDS @ 1500m/s the difference is 2/3.

    In any event this always leads to erosion on the penetrator and added stress. As long as the projectile is penetrating the armor , the projectile is 'supported' by the crater its generating. But the moment it clears the rear of the plate, the projectile is no longer supported and the residual stress waves will strain the projectile, while the velocity of the tip tries to equalise with the tail.

    If the projectile is strong steel [not hard], this will mean a small amount of the penetrator will be lost, but if the penetrator is some brittle material like tungsten carbide , it will shatter.

    With the mesh the penetrator takes the path of least resistance and yaws severly. Again its been shown experimentally that even strong steel penetrators will break in two if they experience lateral velocities of even 70m/s.Under these stresses, WC will shatter into small pieces.

    A whole line of modern armors are based on this concept [On LEopard tank].

  17. Originally posted by Andreas:

    [QB]Paul, hi. I realised I did not put info about the probable strength of 13. PD up here after I read your post.

    Strength return from 1st August 1944 (numbers in brackets are TO&E)

    </font>

    • Panzer III 5 (10)</font>
    • Panzer IV 36 (81)</font>
    • Panzer V Panther 0 (79)</font>
    • SPW, AC, FOO Tanks 75 (264)</font>
    • SP AT 17 (31)
      Artillery</font>
    • Hummel 7 (8)</font>
    • Wespe 16 (16)</font>
    • towed 41 (43)</font>
    • towed heavy AT 41 (71)</font>
    • MG 69 (1,104)
      Vehicles</font>
    • Trucks 697 (1,637)</font>
    • small vehicles 240 (677)
      </font>

    QB]

    I like these figures they look similar to figures for the 1SS Pz Div in 1943....It seems that the division had anywhere from 180-230 tanks and Assault guns. However at no time during that year where they able to field more than half that number . Often only 70-80 tanks were available for offensive opps and after several weeks this was down to ~ 30-40 operational tanks.

    To me, the Germans should have invested more in ARV and repair crews so they could rely on more than 1/4 to 1/2 of there tank force being operational!

  18. Originally posted by Andreas:

    [QB]Paul, hi. I realised I did not put info about the probable strength of 13. PD up here after I read your post.

    Strength return from 1st August 1944 (numbers in brackets are TO&E)

    </font>

    • Panzer III 5 (10)</font>
    • Panzer IV 36 (81)</font>
    • Panzer V Panther 0 (79)</font>
    • SPW, AC, FOO Tanks 75 (264)</font>
    • SP AT 17 (31)
      Artillery</font>
    • Hummel 7 (8)</font>
    • Wespe 16 (16)</font>
    • towed 41 (43)</font>
    • towed heavy AT 41 (71)</font>
    • MG 69 (1,104)
      Vehicles</font>
    • Trucks 697 (1,637)</font>
    • small vehicles 240 (677)
      </font>

    QB]

    I like these figures they look similar to figures for the 1SS Pz Div in 1943....It seems that the division had anywhere from 180-230 tanks and Assault guns. However at no time during that year where they able to field more than half that number . Often only 70-80 tanks were available for offensive opps and after several weeks this was down to ~ 30-40 operational tanks.

    To me, the Germans should have invested more in ARV and repair crews so they could rely on more than 1/4 to 1/2 of there tank force being operational!

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